Q 101-P A ball is thrown down vertically... [FREE SOLUTION]

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Chapter 2: Q 101-P (page 39)

A ball is thrown down vertically with an initial speed of from a height of . (a) What is its speed just before it strikes the ground? (b) How long does the ball take to reach the ground? What would be the answers to (c) part a and (d) part b if the ball were thrown upward from the same height and with the same initial speed? Before solving any equations, decide whether the answers to (c) and (d) should be greater than, less than, or the same as in (a) and (b).

Short Answer

Expert verified

The speed of the ball just before it strikes the ground is

The time taken by the ball to reach the ground is
The speed of the ball just before it strikes the ground,
The speed of the ball just before it strikes the ground, if the ball were thrown upward from the same height and with the same initial speed, will be the same as it thrown downward.
The time taken by the ball to reach the ground if the ball were thrown upward from the same height and with the same initial speed is which is greater that in part (b).

Step by step solution

To understand kinematic equations of motion

Kinematic equations of motion are the set of equations that describe the motion of an object with constant acceleration. In this case the initial velocity and initial displacement has been given. Using these quantities, the kinematic equations can be constructed to find the velocity and time of the ball.

The expression for the kinematic equations of motion are given as follows:

$v = v_{0} + a t$ 鈥� (i)

$v^{2} = {v_{0}}^{2} + 2 a y$

鈥� (ii)

Here, $v_{0}$ is the initial velocity, $t$ is the time, and $a$ is the acceleration.

(a) Determination of the speed just before ball strikes the ground when thrown vertically down

When the ball is thrown vertically down from height h, the acceleration on the ball is positive, .

Therefore, from equation (ii),

$v = {v_{0}}^{2} + 2 g h$

$v = \sqrt{{v_{0}}^{2}} + 2 g h$

Thus, the speed of ball just before it strikes the ground is $v = \sqrt{{v_{0}}^{2}} + 2 g h$ .

(b) Determination of the time taken by ball to reach the ground when thrown vertically down.

When the ball is thrown vertically down from height h, the acceleration on the ball is positive, $a = g$ .

Therefore, from equation (i),

$v = v_{0} + g t$

$t = \frac{- v_{0} + v}{g}$

Substitute the value of $v$ from part (a).

role="math" localid="1650518721505" $t = \frac{- v_{0} + \sqrt{{v_{0}}^{2} + 2 g h}}{g}$

Thus, the time taken by the ball to reach the ground is $t = \frac{- v_{0} + \sqrt{{v_{0}}^{2} + 2 g h}}{g}$ .

(c) Determination of the speed of ball just before it strikes the ground when thrown upward from the same height and with the same initial speed

If the ball would be thrown upward from the same height and with the same initial speed, after some time, it will return to the same height with the same initial speed. So, the speed of the ball just before it strikes the ground will be the same as in (a).

(d) Determination of the time taken by ball to reach the ground when thrown upward from the same height and with the same initial speed

When the ball is thrown vertically up from height h, the acceleration on the ball is positive, $a = g$ and the initial velocity is negative.

Therefore, from equation (i),

$v = - v_{0} + g t$

$t = \frac{v_{0} + v}{g}$

Substitute the value of $v$ from part (a).

$t = \frac{v_{0} + \sqrt{{v_{0}}^{2}} + 2 g h}{g}$

Thus, the time taken by the ball to reach the ground is $t = \frac{v_{0} + \sqrt{{v_{0}}^{2} + 2 g h}}{g}$ which is greater than the time in part (a).