Answer to Question #268878 in Mechanics | Relativity for efe

Question

A baseball is hit with a speed of 29.0 m/s

m/s at an angle of 43.0 

∘ . It lands on the flat roof of a 14.0 m

m -tall nearby building.

Part A

If the ball was hit when it was 1.0 m

m above the ground, what horizontal distance does it travel before it lands on the building?

Answer 1

Explanations & Calculations

• The vertical height it travels is $\small 14.0\,m-1.0\,m =13.0\,m$
• Apply $\small s = ut + \frac{1}{2}at^2$ for the vertical motion to calculate the time it took for the travel.

$\qquad\qquad \begin{aligned} \small \uparrow 13 &=\small (29\sin43)t+\frac{1}{2}(-9 .8ms^{-2})t^2\\ \small 4.9t^2-19.8t+13&=\small0\\ \small x&=\small \begin{cases} \small3.2\\ \small 0.8 \end{cases} \end{aligned}$

• This means the possible time it takes to travel a $\small 13\,m$ vertical distance, the less time for the motion way up while the longer time for the way down.
• Since the ball lands on the horizontal roof, it should occur during its returning/ downward motion. Hence the tie it took is $\small 3.2\,s$
• Then, applying $\small s =ut+\frac{1}{2}(0)t^2$ for the horizontal motion, the horizontal distance can be found.

$\qquad\qquad \begin{aligned} \small \to s&=\small ut\\ &=\small (29\cos43)\times 3.2\,s\\ &=\small 67.9\,m \end{aligned}$