Question

A. 13
A robot standing on a cliff shoots a ball upwards with an initial speed of 30 ms.
What is the height of the cliff if the ball reaches the bottom of the cliff 8 sec after the
de of
(Take g=10 ms-2
B. 45m
25m
shoot?
C, 80m
D. 145m​

Answer 1

Answer:

145

Explanation:

Let us assume the height of tower as x and its case as case 1

so when it fires:

u=30

a=(-10)

v=0

so applying 3rd eq of motion-

v2=u2+2as

0=30^2-20s

s=30^2/20

 =45

so now if we calculate the time of reaching the topmost height-

s=ut+1/2at2

45=30t-5t2

9=6t+t2

t2-6t+9

on factorizing:

t=3

now taking the scene of falling-

t=8-3=5

u=0

s=45+x

a=10

s=ut +1/2 at2

45+x=5^3

x=625-45=580

so option d is correct as it is closest