Question

A small, spring-loaded cannon launches a tennis ball from level ground with an initial speed vi at an angle θi with the horizontal. The tennis ball lands a horizontal distance R from its launch point. The highest point the tennis ball reaches during its flight is a distance R 12 above the ground. In terms of R and g, find the following. (You may ignore air resistance.)(a) the time interval during which the tennis ball is in motion(b) the tennis ball's speed at the peak of its path(c) the initial vertical component of its velocity

Answer 1

Answer:

a. $T\ =\ \dfrac{2v_isin\theta_i}{g}$

b. $v_icos\theta_i$

c. $v_1sin\theta_i$

Explanation:

Given,

*initial velocity of the ball = $v_i$

*angle of projection = \theta_i

Horizontal component of the initial velocity of the ball = $v_x\ =\ v_icos\theta_i$

vertical initial component of the initial velocity of the ball = $v_y\ =\ v_isin\theta_i$

part a

From the kinematics,

In the y-direction motion,

total vertical displacement of the ball during the whole motion is zero.

Ball is moving under the gravitational acceleration, therefore the acceleration of the ball = -g, because gravitational acceleration always acts in the downward direction,

Let t be the time of flight of the whole motion,

$y\ =\ v_yt\ -\ \dfrac{1}{2}gt^2\\\Rightarrow 0\ =\ v_isin\theta_i t\ -\ \dfrac{1}{2}gt^2\\\Rightarrow t\ =\ \dfrac{2v_isin\theta_i}{g}$

part b.

At the peak of the path of the ball, the vertical component of the velocity of the ball becomes zero, only horizontal component of the velocity acts on the ball is equal to = $v_x\ =\ v_icos\theta_i$

part c.

Initial vertical component of the velocity of the ball = $v_y\ =\ v_isin\theta_i$