Serving at a speed of 170 km/h, a tennis player hits the ball at a height of 2.5 m and an angle θ below the horizontal. The baseline is 11.9 m from the net, which is 0.91 m high. What is the angle θ such that the ball just crosses the net? Will the ball land in the service box, whose service line is 6.40 m from the net?
Solution:
Note: The publication of the solution to this problem is on its way. Sorry for the inconvenience.
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
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