A boy stands at 78.4 metre from a building and throws a ball which just enters a window 39.2 metre above the ground. Calculate the velocity of projection of ball?
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According to question
R = 2 × 78.4 m
H = 39.2 m
Relationship between R and H is
H = R Tanx / 4
Then,
Tanx = 4H / R
Tanx = (4 × 39.2) / (2 × 78.4)
Tanx = 1
x = 45°
Now,
H = u^2 Sin^2x / (2g)
u = sqrt(2gH) / Sinx
u = sqrt(2 × 9.8 × 39.2) / Sin45
u = sqrt(2 × 9.8 × 39.2) / (1/sqrt(2))
u = 39.2 m/s
Velocity of projection of ball is 39.2 m/s
R = 2 × 78.4 m
H = 39.2 m
Relationship between R and H is
H = R Tanx / 4
Then,
Tanx = 4H / R
Tanx = (4 × 39.2) / (2 × 78.4)
Tanx = 1
x = 45°
Now,
H = u^2 Sin^2x / (2g)
u = sqrt(2gH) / Sinx
u = sqrt(2 × 9.8 × 39.2) / Sin45
u = sqrt(2 × 9.8 × 39.2) / (1/sqrt(2))
u = 39.2 m/s
Velocity of projection of ball is 39.2 m/s
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