Question

An object is made to fall from the top of a tower of height 78.4m
1) Find the distance travelled by it in the first second.
2) The time after which it strikes the ground.
3) The velocity with which it strikes the ground.
(take g = 9.8 ms-1)

Answer 1

Answer:

(1) 4.9m

(2) 4sec

(3) 39.2m/s

Explanation:

(1) from 2nd equation of motion

s=ut+1/2gt^2

so

t=1s and g=9.8m/s^2 initial velocity u=0m/s because object is not thrown it just fall

s=0×1+1/2×9.8×1^2

s=4.9m

(2)

by second equation of motion

s=ut+1/2gt^2

and

u=0m/s s=78.4m(given)

so

78.4=1/2×9.8×t^2

t=sqrt{78.4×2/9.8}

t=4s

(3)

by 3rd equation of motion

v^2=u^2+2gs

u=0m/s s=78.4m(given)

so

v=sqrt{2×9.8×78.4}

v=39.2m/s