Question

A ball thrown vertically upwards from the
top of a tower with a speed of 40 m/s returns
back to the ground level in 10 s. The height
of the tower is
A.50
b.100
c.200
d.300​

Answer 1

Answer:

200

Explanation:

Given , initial velocity ( u)= 40m/s

final velocity (v) will be zero

time = 10s

So, acceleration = change in the velocity / time

= 0-40/10 = -4m/s^2

(-) ve sign shows that the object is retarded or negatively accelerated.

Now, by using 2nd equation of motion,

s= ut + 1/2at^2

s= 40*10+1/2*(-4)*100

400-200 = 200m .Ans

Hence the height of tower is 200m.

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