Question

A boady is thrown up with a velocity 50m/s.Calculate the maximum height reached and the time taken to reach the maximum height.(a=10m/s²) ​

Answer 1

Answer:

The final velocity of the body thrown = 50 m/s

The initial velocity of the body = 0 m/s

The time taken to reach maximum height = t s

The maximum height reached by the body = s m

The acceleration due to gravity = 10 m/s²

Using equations of motion :

v² - u² = 2as

⇒ 50² - 0² = 2×10×s

⇒ 2500 = 20s

⇒ s = 2500/20

⇒ s = 250/2

⇒ s = 125 m

Hence, the maximum height reached by the body is 125 m.

s = ut + ½at²

⇒ 125 = 0×t + ½×10×t²

⇒ 125 = ½×10×t²

⇒ 125 = 5t²

⇒ t² = 125/5

⇒ t² = 25

⇒ t = √25

⇒ t = 5 s

Hence, the time taken to reach the maximum height is 5 s.

Answer 2

Answer:

heigth=2480,time=5sec

Explanation:

Given,

         v=0m/s

         u=50m/s

         s=h=?

         t=?

         a=-g=10m/s

we know,

                v^2=u^2-2gh

                0^2=50^2-2*10*h

                      =2500-20*h

                      =2480-h

                    h=2480

now,

       v=u-gt

       0=50-10t

       10t=50

           t=5sec