Question

a. With what minimum speed must you toss a $100g$ball straight up to just touch the $10-m$-high roof of the gymnasium if you release the ball $1.5m$above the ground? Solve this problem using energy.

b. With what speed does the ball hit the ground?

Short answer

(a) Therefore, the minimum required speed is $13m/s$.

(b) When the ball strikes the ground, its height is zero. Therefore,$14m/s$.

Step-by-step solution

Introduction

When an object is in free fall, the sum of its initial kinetic and potential energy equals the sum of its ultimate kinetic and potential energy.

$K_i+U_i=K_{\mathrm{f}}+U_{\mathrm{f}}$

Here, $K_i$is the initial kinetic energy, $U_i$is the initial potential energy, $K_{\mathrm{f}}$is the final kinetic energy, and $U_f$is the final kinetic energy.

Here, $K_i$is the initial kinetic energy, $U_i$is the initial potential energy, $K_{\mathrm{f}}$is the final kinetic energy, and $U_f$is the final kinetic energy.

Explanation (a)

This is a free-fall dilemma. During the ball's joumey, the sum of its gravitational potential and kinetic energy remains constant. Therefore,

$K_{\mathrm{i}}+U_i=K_{\mathrm{r}}+U_{\mathrm{f}}$$\frac{1}{2}m{v}_0^2+mg{y}_e=\frac{1}{2}m{v}_1^2+mg{y}_1$$\frac{1}{2}m{v}_0^2-\frac{1}{2}m{v}_1^2=mg{y}_1-mg{y}_0$Here $m$is mass of the ball, $y_0$is the initial height,$y_1$ is the final maximum height, $v_0$is the initia! speed of the ball and $v_1$is the final speed of the ball when it reaches the maximum height. Speed of the ball at the maximum height is zero. Therefore, $\frac{1}{2}m{v}_0^2=mg{y}_1-mg{y}_0$, $v_0^2=2g\left(y_1-y_0\right)$

Substitute $9.8\mathrm{m}/{\mathrm{s}}^2$for $g,10\mathrm{m}$for $y_1$and $1.5\mathrm{m}$for $y_0$.

$v_0=\sqrt{2\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)(10\mathrm{m}-1.5\mathrm{m})}$

$=13\mathrm{m}/\mathrm{s}$

Therefore, the minimum required speed is $13\mathrm{m}/\mathrm{s}$.

Explanation (b)

The ball's total gravitational potential and kinetic energy remain constant throughout its travel. Therefore,

$K_{\mathrm{i}}+U_{\mathrm{i}}=K_{\mathrm{f}}+U_{\mathrm{f}}$

$\frac{1}{2}m{v}_0^2+mg{y}_0=\frac{1}{2}m{v}_2^2+mg{y}_2$

Here,$y_0$is the initial height, $y_2$is the height when the ball hits the ground,$v_0$is the initial speed of the ball, and $v_2$is the final speed of the ball when it hits the ground.

Height is zero when the ball hits the ground. Therefore,

$\frac{1}{2}m{v}_0^2+mg{y}_0=\frac{1}{2}m{v}_2^2$

$m{v}_2^2=m{v}_0^2+2mg{y}_0$

$v_2^2=v_0^2+2g{y}_0$

$v_2=\sqrt{v_0^2+2g{y}_0}$

Substitute $13\mathrm{m}/\mathrm{s}$for $v_0,1.5\mathrm{m}$for $y_0$and $9.8\mathrm{m}/{\mathrm{s}}^2$for $g$.

$v_2=\sqrt{(13\mathrm{m}/\mathrm{s}{)}^2+2\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)(1.5\mathrm{m})}$

$=14\mathrm{m}/\mathrm{s}$