The world long jump record is 8.95 m (Mike Powell, USA, 1991). Treated as a projectile, what is the maximum range obtainable by a person if he has a take-off speed of 9.5 m/s? State your assumptions.
Solution:
We are required to solve for the maximum distance. To do this, we can use the formula for the range of a projectile motion. However, we need the following assumptions:
The jumper leaves the ground in a 45° angle from the horizontal, for maximum horizontal displacement.
The jumper is on level ground, and the motion started from the ground.
In the standing broad jump, one squats and then pushes off with the legs to see how far one can jump. Suppose the extension of the legs from the crouch position is 0.600 m and the acceleration achieved from this position is 1.25 times the acceleration due to gravity, g . How far can they jump? State your assumptions. (Increased range can be achieved by swinging the arms in the direction of the jump.)
Solution:
We are required to solve for the distance in a standing broad jump. To do this, we can use the formula for the range of a projectile motion. However, we need the following assumptions:
The jumper leaves the ground in a 45° angle from the horizontal, for maximum horizontal displacement.
An arrow is shot from a height of 1.5 m toward a cliff of height H . It is shot with a velocity of 30 m/s at an angle of 60º above the horizontal. It lands on the top edge of the cliff 4.0 s later. (a) What is the height of the cliff? (b) What is the maximum height reached by the arrow along its trajectory? (c) What is the arrow’s impact speed just before hitting the cliff?
Solution:
Consider the following illustration:
Part A
We are required to solve for the value of H. We shall use the formula
Since we know that the horizontal component of the velocity does not change along the entire flight, we can equate the initial and final horizontal velocities. That is
The cannon on a battleship can fire a shell a maximum distance of 32.0 km. (a) Calculate the initial velocity of the shell. (b) What maximum height does it reach? (At its highest, the shell is above 60% of the atmosphere—but air resistance is not really negligible as assumed to make this problem easier.) (c) The ocean is not flat, because the Earth is curved. Assume that the radius of the Earth is 6.37×103 km . How many meters lower will its surface be 32.0 km from the ship along a horizontal line parallel to the surface at the ship? Does your answer imply that error introduced by the assumption of a flat Earth in projectile motion is significant here?
Solution:
Part A
We are given the range of the projectile motion. The range is 32.0 km. We also know that for the projectile to reach its maximum distance, it should be fired at 45°. So from the formula of range,
We are solving for the maximum height here, which happened at the mid-flight of the projectile. The vertical velocity at this point is zero. Considering all this, the formula for the maximum height is derived to be
A right triangle is formed with the legs, the horizontal distance and the radius of the earth, and the hypotenuse is the sum of the radius of the earth and the distance d, which is the unknown in this problem. Using Pythagorean Theorem, and converting all units to meters, we have
Verify the ranges for the projectiles in Figure 3.40(a) for θ=45º and the given initial velocities.
Solution:
To verify the given values in the figure, we need to solve for individual ranges for the given initial velocities. To do this, we shall use the formula
A rugby player passes the ball 7.00 m across the field, where it is caught at the same height as it left his hand. (a) At what angle was the ball thrown if its initial speed was 12.0 m/s, assuming that the smaller of the two possible angles was used? (b) What other angle gives the same range, and why would it not be used? (c) How long did this pass take?
Solution:
To illustrate the problem, consider the following figure:
Part A
We are given the 7-meter range, R, and the initial velocity, vo, of the projectile. We have R=7.0 m, and vo=12.0 m/s. To solve for the angle of the initial velocity, we will use the formula for range
This angle is not used as often, because the time of flight will be longer. In rugby that means the defense would have a greater time to get into position to knock down or intercept the pass that has the larger angle of release.
Part C
We can use the x-component of the motion to solve for the time of flight.
\Delta \text{x}=\text{v}_\text{x}\text{t}
We need the horizontal component of the velocity. We should be able to solve for the component since we are already given the initial velocity and the angle.
An archer shoots an arrow at a 75.0 m distant target; the bull’s-eye of the target is at same height as the release height of the arrow. (a) At what angle must the arrow be released to hit the bull’s-eye if its initial speed is 35.0 m/s? In this part of the problem, explicitly show how you follow the steps involved in solving projectile motion problems. (b) There is a large tree halfway between the archer and the target with an overhanging horizontal branch 3.50 m above the release height of the arrow. Will the arrow go over or under the branch?
Solution:
To illustrate the problem, consider the following figure:
Part A
We are given the range of 75-meter range, R, and the initial velocity, vo, of the projectile. We have R=75.0 m, and vo=35.0 m/s. To solve for the angle of the initial velocity, we will use the formula for range
We know that halfway, the maximum height of the projectile occurs. Also at this instant, the vertical velocity is zero. We can solve for the maximum height and compare it with the given height of 3.50 meters.
The maximum height can be computed using the formula
To compute for the maximum height, we need the initial vertical velocity, voy. Since we know the magnitude and direction of the initial velocity, we have
(a) A daredevil is attempting to jump his motorcycle over a line of buses parked end to end by driving up a 32º ramp at a speed of 40.0 m/s (144 km/h) . How many buses can he clear if the top of the takeoff ramp is at the same height as the bus tops and the buses are 20.0 m long? (b) Discuss what your answer implies about the margin of error in this act—that is, consider how much greater the range is than the horizontal distance he must travel to miss the end of the last bus. (Neglect air resistance.)
Solution:
To illustrate the problem, consider the following figure:
Part A
To determine the number of buses that the daredevil can clear, we will divide the range of the projectile path by 20 m, the length of 1 bus. That is
\text{no. of bus}=\frac{\text{Range}}{\text{bus length}}
\begin{align*}
\text{no. of buses} & =\frac{146.7\:\text{m}}{20\:\text{m}} \\
\text{no. of buses} & =7.34\:\text{buses} \\
\text{no. of buses} & =7\:\text{buses}
\qquad \qquad{\color{DarkOrange} \left( \text{Answer} \right)} \\
\end{align*}
Therefore, he can only clear 7 buses.
Part B
He clears the last bus by 6.7 m, which seems to be a large margin of error, but since we neglected air resistance, it really isn’t that much room for error.
A ball is thrown horizontally from the top of a 60.0-m building and lands 100.0 m from the base of the building. Ignore air resistance. (a) How long is the ball in the air? (b) What must have been the initial horizontal component of the velocity? (c) What is the vertical component of the velocity just before the ball hits the ground? (d) What is the velocity (including both the horizontal and vertical components) of the ball just before it hits the ground?
Solution:
To illustrate the problem, consider the following figure:
Part A
The problem states that the initial velocity is horizontal, this means that the initial vertical velocity is zero. We are also given the height of the building (which is a downward displacement), so we can solve for the time of flight using the formula y=voyt+1/2at2. That is,
To solve for the velocity as the ball hits the ground, we shall consider two points: (1) at the beginning of the flight, and (2) when the ball hits the ground.
We know that the initial velocity, voy, is zero. To solve for the final velocity, we will use the formula \text{v}_{\text{f}}=\text{v}_{\text{o}}+\text{at}.
A ball is kicked with an initial velocity of 16 m/s in the horizontal direction and 12 m/s in the vertical direction. (a) At what speed does the ball hit the ground? (b) For how long does the ball remain in the air? (c)What maximum height is attained by the ball?
Solution:
To illustrate the problem, consider the following figure:
Part A
Since the starting position has the same elevation as when it hits the ground, the speeds at these points are the same. The final speed is computed by solving the resultant of the horizontal and vertical velocities. That is
Consider the two points: (1) the starting point and (2) the highest point.
We know that at the highest point, the vertical velocity is zero. We also know that the total time of the flight is twice the time from the beginning to the top.
So, we shall use the formula \text{t}=\frac{\text{v}_{\text{f}}-\text{v}_{\text{o}}}{\text{a}}.
The maximum height attained can be calculated using the formula \left(\text{v}_{\text{f}}\right)^2=\left(\text{v}_{\text{o}}\right)^2+2\text{a}\text{y}.
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