A ball is projected from a tower of height 12 metre with speed of 2m/sec in horizontal direction. Find the horizontal displacement of ball before it strikes the ground.(take g = 10m/s²)
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Answer:
Hence, the horizontal displacement of the ball before it strikes the ground = 180 m
Explanation:
Let the horizontal displacement be x.
And vertical height(height of tower) be y = 12m
Use formula
Y = (g×x^2)÷(2u^2)
……(u is initial velocity=2m/s)
Y = (10×12×12)÷2×4
= 1440÷8
= 180m
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The time of fall=√(2×12/g)
Distance along x direction
2×√(2×12/g)= √(96/10) = √ 9.6m
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