Question

A ball thrown upis caught by the thrower after 4 second. How high did it go and with what velocity was it thrown? Hoe far was it below the highest point 3 second after it was thrown?

Answer 1

t= 4s

a=-10m/s^2

To reach max height , t=2s

And at max height , v=0m/s

Using

v=u+at

0=u-(10)(2)

u= 20m/s

Using

V^2-u^2=2as

0-400=-(2)(10)(s)

S= 20m

Now by s=ut+1/2at^2

At t=3s , we have

S=(20)(3)-1/2(10)3

S=60-15

S=45 m

But we require the answer from max height.

So we have 45-20=25m

So initial velocity = 20m/s

Max height =20m

Height at 3s from max height = 25m

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