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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

Respuesta :

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

[tex]V_{y}^2=V_{o_{y}}^2+2gH \\ H= \frac{V_{y}^2-V_{o_{y}}^2}{2g} [/tex]
 
     In the maximum height, the speed is zero, therefore:

[tex]H= \frac{V_{y}^2-V_{o_{y}}^2}{2g} \\ H= \frac{0-29.4^2}{2*9.8} \\ \boxed {H=44.1m}[/tex]

Number 3

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

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