Answer to Question #317198 in Mechanics | Relativity for Quân Jason

Question

1. A boy wants to throw a water balloon from a roof deck at a height of 3m from the street and hit the far side of the street. He throws the ball at 5 m/s at an angle of 45° to the horizontal. If Vboy = 0, what is the maximum height, h, reached by the balloon above the street?

A. h= 5.9 m B. h = 5.1 m C. h = 3.6 m D. h= 4.6 m

2.This time, suppose the boy gets a running start, vboy = 5m/s toward the street. What is the speed of the balloon, V balloon, when it hits the street

Answer 1

Explanations & Calculations

1.

• Apply $\small v^2=u^2+2as$ upwards for the ball's motion.

$\qquad\qquad \begin{aligned} \small 0^2&=\small (5\sin45)^2+2(-9.8\,ms^{-2})h\\ \small h&=\small 0.64\,m \end{aligned}$

• Then from the height from the street is $\small 0.64\,m+3\,m= 3.64\,m(\approx 3.6\,m)$

2.

• Consider the conservation of mechanical energy from the start to end(street is the level of zero potential energy)

$\qquad\qquad \begin{aligned} \small p.e+k.e&=\small p.e+k.e\\ \small mgh+\Sigma\Big(\frac{1}{2}mv^2\Big)&=\small 0+\frac{1}{2}mV^2\\ \small mgh+\frac{1}{2}mv^2+\frac{1}{2}mv_{boy}^2&=\small \frac{1}{2}mV^2\\ \small V&=\small \sqrt{2gh+v^2+v_b^2}\\ &=\small \sqrt{2(9.8)(3)+5^2+5^2}\\ &=\small 10.4\,ms^{-1} \end{aligned}$