Question

A ball is projected vertically up with a speed of 100 m/s. What will be
the maximum height attained, and the time to reach the maximum
height (9=10 m/s2)​

Answer 1

u=100m/s

v=0m/s

g= -10m/s(as the direction is opposite to gravity)

s=?

By v²-u²=2(-g)s,we get-----

0²-(100)²=2(-10)s

-10000=-20s

s=-10000/-20

s=500

Thus,maximum height attained is 500 metres.

Now,t=?

By v=u+(-g)t,we get----

0=100+(-10)t

-100= -10t

t= -100/-10

t=10

Thus,time taken to reach maximum height is 10 seconds.

Answer 2

Answer:

here,

accleration (a)= -10m/s2 (the ball is moving upwards opposite to the gravity)

inital velocity (u)= 100m/s

final velocity (v)= 0m/s

now

a= v-u/t

or, -10 = 0-100/t

or, -10t= -100

or, t = -100/-10

or, t= 10 seconds

again

s= ut+1/2 at^2

 = 100*10 + 1/2 *-10 * 10^2

 = 1000 - 1000/2

 = 1000- 500

  = 500 metres

therefore, the time is 10 seconds and the height is 500 metres

Explanation: