v=u-gt or 0=15-9.81t since final velocity is 0 and ball is going against gravity. Thus t=1.53 sec.
s=ut-.5gt^2=15-.5x9.81x(1.53)^2
s=22.95-11.48=121.47 which is the height the ball will go. From there it falls freely for a height of 11.47+1=12.47 ft.
s=0xt+.5x9.81xt^2 or t=(12.47x2/9.81)=1.59 secs.
When a ball is thrown in the air at 15mps from a height of 1 ft, how long does it take for the ball to hit the ground?
3 answers
PS: s=11.47 which has appeared as 121.47 at one place in 4th line.
since 1 ft = .3 m,
s(t) = .3 + 15t - 4.9t^2
it hits the ground when s=0, so
.3 + 15t - 4.9t^2 = 0
t = 3.08 sec
s(t) = .3 + 15t - 4.9t^2
it hits the ground when s=0, so
.3 + 15t - 4.9t^2 = 0
t = 3.08 sec
1 answer