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

Answer:-

Known Terms:-

u = Initial speed

v = Final speed

a = Acceleration

t = Time

s = Distance

Given:-

Initial speed of the stone, u = 20 m/s

Final speed of the stone, v = 0

Acceleration attain by the stone, a = -9.8 m/s²

Solution:-

For the 1st stone:-

Let S₁ be the distance travelled by the stone in upward direction

From the 3rd equation of motion,

⇒ v² - u² = 2as

⇒ 0 - (20)² = 2 × (-9.8) × S₁

⇒ S = 20.40 m  

Also, $t=\frac{v-u}{a}$

⇒ $t=\frac{v-u}{a}$

⇒ $t=\frac{0-20}{-9.8}$

⇒ $t=2 \: sec$

Now, distance or height = 120.40

t = ???

u = 0

⇒ S = $\frac{1}{2}at^2$

⇒ 120.40 = $\frac{1}{2}(9.8)(t)^2$            [∵ a = g]

⇒ t = 4.95 sec

Total time = t₁ + t = 6.95 sec

For Second Stone:-

⇒ v² - u² = 2gh

⇒ v² - (20)² = 2 × 9.8 × 100

⇒ v² = 20² + (2 × 9.8 ×100)

⇒ v² = 48.59 m/s

Also, $t=\frac{v-u}{a}$

⇒ $t=\frac{v-u}{a}$

⇒ $t=\frac{45.59-20}{9.8}$

⇒ $t=2.92 \: sec$