A 14.7m high vertical wall is situated at a distance of 19.6m from a person.At what angle with the horizontal should an object be thrown by that personat a speed of 19.6 m/s so that it clears the wall?
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1 Answers
Samarth Kashyap
·2009-06-20 19:24:25
v0=vxi + vyj
vx=v0cosθ
vy=v0sinθ
at some instant of time t, x=19.6m and y=14.7m
pls forgive me for the bad diagram
x=vxt
y=vyt - 1/2gt2
subst. for vx and vy.... and solve for θ