Question

A ball is projected vertically with a speed of 50 metre per second find the maximum height ,the time of reach the maximum height and the speed of half of the maximum height ​

Answer 1

Here we can use equation of motion..i.e., V^2=U^2-2gh the main formula is V^2=U^2+2as....negative sign we use because motion is in upward direction and g is acting downward and place of a we use g that is gravitational acceleration....so when ball reaches maximum height final velocity becomes zero and initial velocity=50m/s and g=10m/s^2...so putting all values we get h=125m....for time we can use v=u-gt where v=0,u=50m/s so putting all values we get t=5s and for finding speed of half of the maximum height we will use same equation that we use in first that is v^2=u^2-2gh we have to find v and half height = 62.5 m so putting all values we get....V=35.5m/s...i hope it hepls...