Question

A ball thrown horizontally at vi = 27.0 m/s travels a horizontal distance of d = 53.0 m before hitting the ground. From what height h was the ball thrown? Express your answer numerically in meters. View Available Hint(s) h = nothing m

Answer 1

Answer:

The height from which the ball was thrown is 18.82 m

Explanation:

Given;

horizontal velocity of the ball, vi  = 27.0 m/s

horizontal distance of the ball, d = 53.0 m

Apply kinematic equation, to determine the time taken for the ball to make a horizontal distance of 53.0 m.

d = vt

t = d/v

t = 53/27

t = 1.96 seconds

This time is equal to the time the ball spent in air before hitting the ground.

The vertical distance at this time, is the height from which the ball was thrown, and it is calculated as;

h = vt + ¹/₂gt²

v is vertical velocity,  = 0

g is acceleration due to gravity

h = ¹/₂ x 9.8 x (1.96)²

h = 18.82 m

Therefore, the height from which the ball was thrown is 18.82 m

Answer 2

The height from which the ball is thrown , is 18.82m.

The time taken by ball to hit ground is,

                         $time=\frac{distance}{velocity} \\\\time=\frac{53}{27}=1.96s$

The vertical distance at this time, is the height from which the ball was thrown, and it is calculated by equation shown below,

          $h = vt + \frac{1}{2} gt^{2}$ , where v is vertical velocity.

Vertical velocity, $v=0$

          $h=\frac{1}{2} gt^{2}= \frac{1}{2} (9.8)(1.96)^{2}\\\\h=\frac{37.65}{2}=18.82m$

Therefore, the height from which the ball was thrown is 18.82m.

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