Question

a ball is projected upward from the top of the building with a velocity 30m/s making an angle 30 degree with the horizontal after how much time the ball strikes the ground if the building is 100m high.

Answer 1

The horizontal components

Initial velocity = Ux = ucos∅ = 30cos30° = 30×√3/2 = 15√3 m/s

acceleration = Ax = 0 m/s² (because acceleration due to gravity act always below on y axis so it make 90° angle to x axis i.e horizontal so according to the resolution of the vector then horizontal acceleration becomes -gcos90° in which cos90° is equal to the 0 so acceleration in horizontal becomes -g(0) = 0 m/s²)

final velocity = Vx = Ux = 15√3m/s (because here acceleration is equal to the 0 hence no change in velocity ) .
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The vertical components
initial velocity = Uy = usin∅ = 30sin30° = 30×1/2 = 15 m/s .

distance covered = Sy = -100m (because below of y axis)

acceleration due to gravity = Ay = -g = -10m/s² (or take -9.8 m/s²)

time taken to reach the ground = t
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therefore,
S = ut + at²/2

Sy = Uyt + Ayt²/2

-100 = 15t +(-10)t²/2

-100 = 15t -5t²

5t² -15t -100 = 0

5(t²-3t -20) = 0

t²-3t -20 = 0

by applying the quadratic equation

t = -b±√b²-4ac/2a

t = -(-3)±√(-3)³-4(1)(-20)/2(1)

t = 3±√89/2

t = 3-√89/2 Or t = 3+√89/2

here time is never be negative

t = 3+√89/2

t = 3+9.5/2

t = 12.5/2

t = 6.25 s

so time taken to reach the ground is approximately 6.25 second .
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In 6.25 second horizontal range become

S = ut + at²/2

Sx = Uxt + Axt²/2

Sx = 15√3(6.25) +0

Sx = 15(1.73)(6.25)

Sx = 162.19 meter
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