A basketball player achieves a hang time of 0.827 s in dunking the ball. What ve

Question

A basketball player achieves a hang time of 0.827 s in dunking the ball. What vertical height will he attain? The acceleration of gravity is 9.8 m/s~2 . Answer in units of m A bullet is fired straight up from a gun with a muzzle velocity of 136 m/s. Neglecting air resistance, what will be its displacement after 4.6 s? The acceleration of gravity is 9.8 m/s2 . Answer in units of m Janet jumps off a high diving platform with a horizontal velocity of 1.57 m/s and lands in the water 2.9 s later. How high is the platform? The acceleration of gravity is 9.8 m/s2 . Answer in units of m

Explanation / Answer

1. Taking upward motion as positive, s = 0, t = 0.827 a = -9.8 u = ? s = ut + 0.5at^2 0 = 0.827u + (0.5)(-9.8)(0.827)^2 u = 0.5(9.8)(0.827) = 4.0523 considering displacement to max ht, u = 4.0523 a = -9.8 v = 0 s' = ? v^2 = u^2 + 2as' 0 = 4.0523^2 + 2(-9.8)s' s' = 4.0523^2 / 2(9.8) = 0.838m (3sf) vertical height = 0.838m 2. Taking upward motion as positive, u = 136 a = -9.8 t = 4.6 s = ? s = ut + 0.5at^2 s = 136(4.6) + 0.5(-9.8)(4.6)^2 = 522m (3sf) Displacement = 522m upwards 3. Taking downward motion as positive, u = 0 (jumps with horizontal velocity, ie zero vertical velocity) a = 9.8 t = 2.9 s = ? s = ut + 0.5at^2 s = 0(2.9) + 0.5(9.8)(2.9)^2 = 41.2 (3sf) Height of platform = 41.2m

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