Horizontal problem:
u = horizontal speed = 27 cos 42 forever
so horizontal distance is u T
to find T do vertical problem
vertical problem
Vi = 27 sin 42
Hi = 1.3
h = 14 at t = T
h = Hi + Vi t - 4.9 t^2
so at h = 14 and t = T
14 = 1.3 + 27 sin 42 T - 4.9 T^2
solve quadratic for T
distance = u T
A baseball is hit with a speed of 27.0 m/s at an angle of 41.0 ∘ . It lands on the flat roof of a 14.0 m -tall nearby building. If the ball was hit when it was 1.3 m above the ground, what horizontal distance does it travel before it lands on the building?
1 answer