Question

a ball initially at the top of a inclined hill is allowed to roll down.at the bottom its speed is 4m/s. next the ball is again rolled down the hill but with a initial velocity of 3m/s.how fast is it going when it gets to the bottom?​

Answer 1

Answer:

5m/s

Explanation:

The ball which was initially at rest acquires a speed $\sqrt{2gh}\ m/s$ at the bottom of the inclined plane

⇒√(2gh)= 4

⇒ 2gh=16

⇒gh=8

h= 0.815m

now second ball already has an initial speed of 3m/s.

Therefore, to find its final speed we use

$v^2=u^2-2gh$

$v^2= 3^2+16=9+16=25\\\\v=\sqrt{25} =5 m/s$

Answer 2

When it gets to the bottom, it gets speed of 5m/s

• We have, mgh = 1/2 mv^2

gh=1/2 v^2

• According to the question,

gh = 1/2* 16 = 8

h = 8/g

• For the second part,

mgh + 1/2 mv^2 = 1/2 mV^2

2gh + v^2 = V^2

16 + 9 = V^2

V^2 = 25

• Since speed cannot be negative,

V = 5 m/s