Question

Two stones are projected from the top of a tower 100m high,each with a velocity of 20m/s,one is projected vertically upward and the other vertically downward:

calculate the time each stone takes to reach the ground.with what velocity will it strike the ground.

Answer 1

stone projected vertically upward


height upto which it goes above(h) can be calculated as
v^2=u^2+2aS
v=0{final speed will be zero at the top}, a= negative{against gravity}
u^2=2aS
20×20= 2×10× h
h=20
total height from which it fall in the end= 100+20
=120
height travelled = ut + 1/2gt^2
120= 0+1/2×10t^2
t^2=24
t=2√6sec
v^2=u^2 +2a(total height)
v^2=0+ 2×10×120
v=20√6


stone projected vertically downward

v^2=u^2+2aH
v^2=20×20+ 2×10×100
=400+2000
=2400
v=20√6
v=u+gt
20√6=20+10t
2√6-2=t

Answer 2

Answer:

Explanation:

Solution :-

For the first stone -

Let s' be the distance traveled by the stone in upward direction.

u = 20 m/s

v = 0

a = - 9.8 m/s²

We know that,

⇒ v² - u² = 2as

⇒ (0)² - (20)² = 2 × (- 9.8) × s'

⇒ s' = 20.40 m.

Also,

t' = v - u/a

⇒ t' = 0 - 20/- 9.8

⇒ t' = 2 seconds

Now, distance or height = 20.40 m

t = ?

u = 0

We know that

⇒ s = ut + 1/2 at²

⇒ 20.40 = 1/2 × 9.8t²  (a = g)

⇒ t = 4.95 seconds

Total time = t' + t = 6.95 s

For second stone -

We know that,

⇒ v² - u² = 2gh

⇒ v² - (20)² = 2 × 9.8 × 10

⇒ v² = 48.59 m/s²

Now,

t = v - u/g

⇒ t = 48.59 - 20/9.8

⇒ t = 2.92 seconds.