Chapter 2, Problem 63P

(a)

To determine

The distance covered by the ball in first $3\text{s}$.

Answer to Problem 63P

The distance covered by the ball in first $3\text{s}$ is $44.1\text{m}$.

Explanation of Solution

The acceleration of the ball is $9.80\text{m}/{\text{s}}^{\text{2}}$ and time taken is $3\text{s}$.

Write the expression to calculate the distance covered.

$s=\frac{1}{2}a{t}^2$

Here, s is the distance covered, a is the acceleration of the ball and t is the time taken.

Substitute $9.80\text{m}/{\text{s}}^{\text{2}}$ for a and $3\text{s}$ for t in the above equation to calculate s.

$\begin{array}{c}s=\frac{1}{2}\left(9.80\text{m}/{\text{s}}^{\text{2}}\right){\left(3\text{s}\right)}^2\\ =44.1\text{m}\end{array}$

Conclusion:

Therefore, the distance covered by the ball in first $3\text{s}$ is $44.1\text{m}$.

(b)

To determine

The speed of the ball.

Answer to Problem 63P

The speed of the ball is $7.0\text{m}/\text{s}$.

Explanation of Solution

The distance moved is $2.5\text{m}$ and acceleration of the ball is $9.80\text{m}/{\text{s}}^{\text{2}}$.

Write the expression for the speed of the ball.

$v=\sqrt{2ah}$

Here, v is the speed of the ball and h is the distance moved.

Substitute $2.5\text{m}$ for h and $9.80\text{m}/{\text{s}}^{\text{2}}$ for a in the above equation to calculate v.

$\begin{array}{c}v=\sqrt{2\left(2.5\text{m}\right)\left(9.80\text{m}/{\text{s}}^{\text{2}}\right)}\\ =7.0\text{m}/\text{s}\end{array}$

Conclusion:

Therefore, the speed of the ball is $7.0\text{m}/\text{s}$.

(c)

To determine

The speed of the ball after $3\text{s}$.

Answer to Problem 63P

The speed of the ball after $3\text{s}$ is $29.4\text{m}/\text{s}$.

Explanation of Solution

The initial speed of the ball is $0\text{m}/\text{s}$, the time taken is $3\text{s}$ and acceleration is $9.80\text{m}/{\text{s}}^{\text{2}}$.

Write the expression to calculate the speed of the ball.

$v=u+at$

Here, v is the speed of the ball, u is the initial speed of the ball and t is the time taken.

Substitute $0\text{m}/\text{s}$ for u, $3\text{s}$ for t and $9.80\text{m}/{\text{s}}^{\text{2}}$ for a in the above equation to calculate v.

$\begin{array}{c}v=0\text{m}/\text{s}+9.80\text{m}/{\text{s}}^{\text{2}}\left(3\text{s}\right)\\ =29.4\text{m}/\text{s}\end{array}$

Conclusion:

Thus, the speed of the ball after $3\text{s}$ is $29.4\text{m}/\text{s}$.

(d)

To determine

The position of the ball.

Answer to Problem 63P

The position of the ball is $\text{17}\text{.1}\text{m}$ below the top of the tower.

Explanation of Solution

The initial speed of the ball is $4.80\text{m}/\text{s}$, the time taken is $2.42\text{s}$ and acceleration is $-9.80\text{m}/{\text{s}}^{\text{2}}$.

Write the expression for the position of the ball after $2.42\text{s}$.

$y=ut+\frac{1}{2}a{t}^2$

Here, y is the position of the ball, u is the initial speed of the ball, t is the time taken.

Substitute $4.80\text{m}/\text{s}$ for u, $2.42\text{s}$ for t and $9.80\text{m}/{\text{s}}^{\text{2}}$ for a in the above equation to calculate y.

$\begin{array}{c}y=\left(4.80\text{m}/\text{s}\right)2.42\text{s}+\frac{1}{2}\left(-9.80\text{m}/{\text{s}}^{\text{2}}\right){\left(2.42\text{s}\right)}^2\\ =11.62\text{m}-28.70\text{m}\\ \text{=}-\text{17}\text{.1}\text{m}\end{array}$

Thus, the ball is $\text{17}\text{.1}\text{m}$ below the top of the tower.

Conclusion:

Thus, the position of the ball is $\text{17}\text{.1}\text{m}$ below the top of the tower.