Question

A ball is thrown up with a speed of 15 m/s find the highest level it will reach (g=9.8m/s)

Answer 1

Given:-

• Initial velocity ,u = 15m/s

• Final velocity, v = 0m/s

• Acceleration due to gravity ,g= 9.8m/s

To Find:-

• Height level attained by ball ,h

Solution:-

By using 3rd equation of motion

• v² = u² +2ah

Substitute the value we get

→ 0² = 15² + 2×(-9.8) ×h

→ 0 = 225 + (-19.2)×h

→ -225 = -19.2×h

→ h = -225/-19.2

→ h = 225/19.2

→ h = 11.71 m

∴ The height attained by the ball is 11.71 m.

Answer 2

Question

A ball is thrown out with a speed of 15 metre per second find the highest level it will reach (g=9•8m/s) ?

Answer

What is given here ?

• T = 1 sec
• acceleration due to gravity =9.8m/s
• the velocity is zero metre per sec

what we have to find ?

• we how to find Here the height of the ball

How to find it ?

• we have to first find out the initial velocity then we can find the height of the ball will reach .

Solution

let's find out the initial velocity ,

v = u + at

=> 0 = u+1 (9.8)

=> 0=u -9.8

=>-u=-9.8

=> u=9.8

so the intial velocity is 9.8 m / s.

now let us find out the height of the ball will reach by using the third equation of motion ,

=> v^2=u^2+2as

=> 0^2=9.8^2+(-19.6)×s

=> 0=96.04 +(-19.6)×s

=> 0=-96.04=(-19.6)×s

=> s=-96.04/19.6

=>s=4.9 m

so the ball will reach the height of 4.9 m