Question

A ball is dropped from a tower it reaches the ground in 10 seconds. Calculate the height of the tower and the velocity with which it hits the ground.​

Answer 1

Answer:

Data: t=10 seconds, height=?, Velocity =?

Solution: h=5t^2=5(10)^2=5(100)=500m

V=10t=10(10)=100.

Or

V=√2gh=√2×9.8×500=√9800=100.

Or

S=vit+1/2at^2

S=0+1/2(9.8)(10)^2

S=1/2(980)=490=500m.

Answer 2

Given:

• Ball is dropped.
• It takes 10 seconds to reach ground.

To find:

• Height of tower?
• Velocity of impact on ground?

Calculation:

Since the ball was dropped, it's initial velocity will be zero. Also, it's acceleration will be equal to g (i.e. 10 m/s²).

Applying EQUATIONS OF KINEMATICS:

$v = u + gt$

$\implies \: v = 0 + 10(10)$

$\implies \: v = 100 \: m {s}^{ - 1}$

Again applying EQUATIONS OF KINEMATICS:

${v}^{2} = {u}^{2} + 2gh$

$\implies \: {100}^{2} = {0}^{2} + 2(10)h$

$\implies \: 10000 = 20h$

$\implies \:h = 500 \: m$

So, height of tower is 500 m & velocity of impact is 100 m/s.