Question

A ball is thrown vertically upward with an initial

velocity of 29.4 meters per second. What is the
maximum height reached by the ball? [Neglect
friction.]
(1) 14.7 m (3) 44.1 m
(2) 29.4 m (4) 88.1 m

Answer 1

Using the Torricelli's Equation in vertically direction, we have:


`V_(y)^2=V_{o_(y)}^2+2gH \\ H= \frac{V_(y)^2-V_{o_(y)}^2}{2g}`

In the maximum height, the speed is zero, therefore:


`H= \frac{V_(y)^2-V_{o_(y)}^2}{2g} \\ H= (0-29.4^2)/(2*9.8) \\ \boxed {H=44.1m}`

Number 3

If you notice any mistake in my english, please let me know, because i am not native.

Answer 2

This involves suvat equations, so the easiest thing to do is to put the suvat down and see what you have:
S - We are looking for this
U - 29.4
V - 0 (as we want the height at which the ball stops)
A - -9.81 (acceleration due to gravity acting in the opposite direction to the velocity of the ball)
T - We don't have this but we don't need it either

An equation which uses these all is v^2=u^2+2as , so:

0=29.4^2+(2*(-9.81)*s)

-29.4^2=-19.62s
29.4^2=19.62s
s=29.4^2/19.62

s=44.1m to 3sf

Therefore the height reached by the ball is 44.1 metres.