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 = $v_y\ =\ v_isin\theta_i$

Answer 2

The tennis ball moves in projectile motion when spring-loaded cannon launches it.

• (a) The time interval during which the tennis ball is in motion is,
•              $T=\dfrac{2v_i\sin\theta}{g}$
• (b) The tennis ball's speed at the peak of its path is $v_i\cos\theta_i$
• (c) The initial vertical component of its velocity $v_i\sin \theta_i$.

What is projectile motion?

Projectile motion is the motion of the body, when it is thrown in the air taking the action of gravity on it.

Given information-

The initial speed of the ball is $v_i$.

The distance traveled by the tennis ball is R.

The angle of projection is $\theta_i$.

• (a) the time interval during which the tennis ball is in motion-

Time interval during which the tennis ball is in motion is called the time of flight. It can be given as,

$T=\dfrac{2u_0\sin\theta}{g}$

As the initial speed of the ball is $v_i$. Thus, time of flight is,

$T=\dfrac{2v_i\sin\theta}{g}$

Hence, the  time interval during which the tennis ball is in motion is,

$T=\dfrac{2v_i\sin\theta}{g}$

• (b) The tennis ball's speed at the peak of its path-

For the projectile motion, t when tennis ball reaches to the highest point, the vertical component of velocity does not acts on it.

Thus the tennis ball's speed at the peak of its path will be equal to the horizontal component of velocity, which is,

$v_p=v_x=v_i\cos\theta_i$

Hence, the tennis ball's speed at the peak of its path is $v_i\cos\theta_i$

• (c) The initial vertical component of its velocity-

The initial vertical component of the velocity is the product of initial velocity to the sine angle of it, which is,

$v_y=v_i\sin \theta_i$

Thus initial vertical component of its velocity $v_i\sin \theta_i$.

Hence,

• (a) The time interval during which the tennis ball is in motion is,

             $T=\dfrac{2v_i\sin\theta}{g}$

• (b) The tennis ball's speed at the peak of its path is $v_i\cos\theta_i$
• (c) The initial vertical component of its velocity $v_i\sin \theta_i$.

Learn more about the projectile motion here;

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