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3 years ago

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A ball is batted straight up into the air and reaches a maxium height 65.6 m (a) How long did it take to reach this height? (b)

What was the pop-up velocity of the ball?

1 answer:

kondaur [170]3 years ago

6 0

Answer:

a) 3.65 seconds

b) 35.87 m/s

Explanation:

s = Displacement = 65.6 m

u = Initial velocity

v = Final velocity

t = Time taken

a = Acceleration due to gravity = 9.81 m/s² (downward direction is taken as positive and upward is taken as negative)

b) Equation of motion

$v^2-u^2=2as\\\Rightarrow 0^2-u^2=2\times -9.81\times 65.6\\\Rightarrow u=\sqrt{2\times 9.81\times 65.6}\\\Rightarrow u=35.87\ m/s$

Initial pop up velocity is 35.87 m/s

a)

$v=u+at\\\Rightarrow t=\frac{v-u}{a}\\\Rightarrow t=\frac{0-35.87}{-9.81}\\\Rightarrow t=3.65\ s$

It took 3.65 seconds to reach this height

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