Question

A ball is thrown vertically upward with velocity of 10 m/s. Calculate the height attained by the ball and the time taken by it to reach the highest point if the acceleration of the ball is -10 m/s2 during itsmotion.​

Answer 1

Answer:

$v = u + at \\ at \: highest \: point \: v = 0 \\ 0 = 10 + ( - 10)t \\ - 10 = - 10t \\ t = 1sec \\ v {}^{2} - u { }^{2} = 2as \\ as \: v \: = 0 \\ - 10 {}^{2} = 2( - 10) \times s \\ - 100 = - 20s \\ s = 5m$

hope it helps u

Answer 2

Answer:

Initial velocity (u) = 10m/s

Final Velocity (v) = 0m/s

Acceleration (a) = -10m/s

a) Let Height = h

Using,

v^2-u^2 = 2ah

0^2 - 10^2 = 2*-10*h

-100/-20 = h

h = 5m

b) Let time taken = t

Using v = u+at

0 = 10+(-10)*t

t =-10/(-10)

t = 1

Therefore, Maximum height achieved = h = 5m

                , Time taken to achieve max height = t = 1 sec