Ball thrown vertically upward with a speed of 272ms from a h

Ball thrown vertically upward with a speed of 27.2m/s from a height of 2.2m. how long does it take to reach its highest point? answer in units of s.

how long does the ball take to hit the ground after it reaches its highest point? answer in units of s.

Solution

The intial velocity of the ball = 27.2 m

At the highest height, the ball\'s final velocity is 0 .

Using the equation of motion, v = u-gt     (1),             where u is initial velocity and v is final velocity and g is the acceleration due to gravity and t is the time taken by the ball to reach the highest point.

Given u= 27.2m/s, v=0 at thighest point, g is assumed to be 9.8m/s^2. Substituting in (1) we get.

0=27.2-9.8t

t=27.2/g = 27.2/9.8= 2.775510204 secs for the ball to reach the highest point .

The height of highest position of the ball = Average velocity * time to reach

=(27.2+0)/2][27.2/9.8)= 27.2^2/(2*9.8) =37.75 meter from the initial height 2.2m. So, height ground =2.2+37.75= 39.95 meter .

Time to reach the ground from the height H is therefore given by

H= (1/2)gt^2, H=39.95.

39.95=(1/2)gt^2

t^2= 2*39.95/9.8

t=2.86 secs.