`S = ut-1/2 g t^2` is a formulat which given the distance S in meters travelled by a ball from the thrower's hadns if it is thrown upwards with an initial velocvity of 14`m//s` after a time of t seconds. G is the acceleration due to the gravity and is `9.8 m//s^2`
How long does it take for the ball to reach a height of 5 meters?

S = ut-1/2 g t^2 is a formulat which given the distance S in meters travelled by a ball from the thrower's hadns if it is thrown upwards with an initial velocvity of u m//s after a time of t seconds. G is the acceleration due to the gravity and is 9.8 m//s^2 If a ball is thrown upwards at 14m//s how high has it gone after 1 second?

A small ball is thrown at an angle 45^(@) to the horizontal with an initial velocity of 2 sqrt2m//s. The magnitude of mean velocity averaged over the first 2 s is [take, acceleration due to gravity, g = 10 m//s ^(2)]

A ball of mass 0.2kg is thrown from a height of 1m and with an initial velocity of sqrt10 m //s at an angle of 45^(@) with the horizontal. Assuming, acceleration due to gravity g =10m//s ^(2), ten modulus of momentum increment during the total time of motion in kg m/s is

If the distance S travelled by a particle in time t is given by s = t^(2) - 2t + 5 then its acceleration is

Find the total energy of a body of 5 kg mass , which is at a height of 10 m from the earth and falling downwards straightly with a velocity of 20 m/s (Take the acceleration due to gravity as 10m//s^(2) )

If the distance s travelled by a particle in time t is given by s=t^2-2t+5 , then its acceleration is