Suppose that the boy in Problem 3.60 throws the ball upward at 60^o above the horizontal, but all else is the same. Repeat parts (a) and (b) of that problem.

Problem 3.60: A boy 12.0 m above the ground in a tree throws a ball for his dog, who is standing right below the tree and starts running the instant the ball is thrown. If the boy throws the ball horizontally at 8.50 m/s, (a) how fast must the dog run to catch the ball just as it reaches the ground, and (b) how far from the tree will the dog catch the ball?

The velocity contains two components, one is a horizontal component and another one is vertical.

According to Newton’s laws of motion,

$\mathrm{s}=\mathrm{ut}+\frac{1}{2}{\mathrm{at}}^2$

Where, s, u, t, and a are displacement, initial velocity, time and acceleration respectively.

The vertical distance, s=12.0 m

The horizontal velocity of the ball, u=8.5 m/s

Angle, $\mathrm{\theta }={60}^{\mathrm{o}}$

The horizontal and vertical velocity components are and respectively. So the horizontal and vertical velocity will be

$\begin{array}{rcl}\mathrm{u}\mathrm{cos\theta }& =& 8.5\mathrm{m}/\mathrm{s}\times \cos {60}^{\mathrm{o}}\\ & =& 4.25\mathrm{m}/\mathrm{s}\end{array}$

$\begin{array}{rcl}\mathrm{u}\mathrm{sin\theta }& =& 8.5\mathrm{m}/\mathrm{s}\times \sin {60}^{\mathrm{o}}\\ & =& 7.36\mathrm{m}/\mathrm{s}\end{array}$

The boy will have to travel at the same speed to reach the same horizontal distance as the ball, so the speed also be the same that is 4.25 m/s.

For the vertical motion, the initial velocity is zero.

$\begin{array}{rcl}\mathrm{s}& =& \mathrm{ut}+\frac{1}{2}{\mathrm{at}}^2\\ 12\mathrm{m}& =& \left(7.36\mathrm{m}/\mathrm{s}\times \mathrm{t}\right)+\frac{1}{2}\left(9.8\mathrm{m}/{\mathrm{s}}^2\times {\mathrm{t}}^2\right)\\ \mathrm{t}& =& 2.48\mathrm{s}\end{array}$

For the horizontal motion, gravity is zero

$\begin{array}{rcl}\mathrm{s}& =& \mathrm{ut}+\frac{1}{2}{\mathrm{at}}^2\\ & =& 4.25\mathrm{m}/\mathrm{s}\times 2.48\mathrm{s}+\frac{1}{2}\times 0\mathrm{m}/{\mathrm{s}}^2\times {\left(2.49\mathrm{s}\right)}^2\\ & =& 10.54\mathrm{m}\end{array}$

Hence, the horizontal distance traveled by the dog is 10.54 m.