A boy throws a ball upwards from the top of a tower at a speed of 12ms-1.on the way down it just misses the tower and falls to the group 50 m below. Find how long the ball remain in the air?
Answers
Answered by
0
first find time taken to reach highest point
u^2=2as. take g as 10m/s^2
s=144/20=7.2
now
time taken to reach that height
s = ut -1/2gt^2
7.2= 12t -5t^2
5t^2-12t+7.2 =0
t=1.2 sec
now time taken to reach ground
since height of tower is 50 m and height reached is 7.2 m so total height = 57.2
time taken to reach
s = 1/2at^2
57.2 =5t^2
t= root 286 take its approx value as 16.9
so totaltime = 1.2 + 16.9 =18.1 sec
u^2=2as. take g as 10m/s^2
s=144/20=7.2
now
time taken to reach that height
s = ut -1/2gt^2
7.2= 12t -5t^2
5t^2-12t+7.2 =0
t=1.2 sec
now time taken to reach ground
since height of tower is 50 m and height reached is 7.2 m so total height = 57.2
time taken to reach
s = 1/2at^2
57.2 =5t^2
t= root 286 take its approx value as 16.9
so totaltime = 1.2 + 16.9 =18.1 sec
Similar questions