Vi = 75 sin 60 = 65 m/s
u = 75 cos 60 = 37.5 m/s
to go 27 m horizontal takes
t = 27/37.5 = .72 seconds
How high will it be (I assume that is the question) ?
v = Vi - 9.8 t
h = 0 + Vi t - 4.9 t^2
h at .72 s = 65(.72) - 4.9 (.72)^2
= 44.25 m high
A ball of mass 0.3 kg, initially at rest, is projected from ground level toward a wall that is 27.0 m away. The ball's velocity the moment it is projected is 75.0 m/s at 60° relative to the horizontal, and the wall is 11.0 m high. During its flight, the ball impacts nothing else and is not subjected to air resistance.
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