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Question

A ball is thrown with a velocity whose horizontal component is 12 ms1 from a point 15 m above the ground and 6 m away from a vertical wall 18.75 m high is such a way to just clear the wall. At what time(in s)will be ball reach the ground? (take g=10 ms2)


Solution

Horizontal component of velocity, ucosθ=12,
Using 2nd equation of motion in horizontal and vertical direction respectively when ball comes to same horizontal level from where its projection started i.e. from A to B,
6=ucosθtt=0.5 s(1)
3.75=usinθt12gt2(2)

Putting the value of t=0.5 s in (2),
154=usinθ×121210(12)2
usinθ=10
Using 2nd equaton of motion in vertical direction from B to C,
15=10T1210T2T22T3=0T=3, 1T=3 s

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