Vi = 30 sin 38
h = Ho + Vi t - 4.9 t^2
h = 0 at ground
so
0 = 0 + Vi t - 4.9 t^2
t (Vi-4.9 t) = 0
t = 0 (it started at ground
t = Vi/4.9 when it hit ground again
max h is when v = 0
v = Vi -9.8 t
0 = Vi -9.8 t
t = Vi/9.8 at top *** this should be half the whole time in the air***
h = 0 +Vi t - 4.9 t^2
using that t which should be half the total t
u = 30 cos 38
d = u t where t is the total time in the air (not the half time you used for part B)
A soccer ball is kicked into the air at an angle of 38 degrees above the horizontal. The initial velocity of the ball is +30.0m/s. How long is the ball in the air?
-What is the maximum height reached by the ball?
-What is the horizontal distance traveled by the ball?
2 answers
3.766 seconds, 25.544 m, 88.886 m,