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.
You must be logged in to post a comment.