Question

An object travels a distance of 40 m during last
two seconds of its fall when it dropped from top
of a tower. The height of the tower is
[g = 10 m/s?]
(1) 85 m
(2) 75 m
(3) 80 m
(4) 45 m​

Answer 1

Answer:

d option 45 m

Explanation:

s= ut+1/2gt^2

Answer 2

Answer:

4) 45m

Explanation:

we are given that distance traveled in last 2 seconds

lts assume the object took time t to travel the distance.

• distance traveled in last 2 seconds = Distance traveled in time t - distance traveled in time (t-2)

⇒ $S_{L2}$ = $S_{t} - S_{t-2}$

so we need to calculate

• Distance traveled in time t

• and distance traveled in time (t-2)

so the total distance traveled in time t ,$S_{t}$ = $ut + \frac{1at^{2} }{2}$

initial velocity, u = 0

acceleration , a = 10 m/s

put these values and we get

$S_{t}$  = $5t^{2}$

and distance traveled in time (t-2), $S_{t-2}$ = $u(t-2) + \frac{1a(t-2)^{2} }{2}$

initial velocity, u = 0

acceleration , a = 10 m/s

put these values and we get

$S_{t-2}$  = $5(t-2)^{2}$

as already mentioned

distance traveled in last 2 seconds = Distance traveled in time t - distance traveled in time (t-2)

⇒ $S_{L2}$ = $S_{t} - S_{t-2}$

$S_{L2}$ = 40 = $5t^{2} - 5(t-2)^{2}$

solving this equation we get

t = 3s

height of tower is Distance traveled in time t = $S_{t}$  = $5t^{2}$

put value of t = 3 we get

$S_{t}$ = 45m