Joe and Bill throw identical balls vertically upward. Joe throws his ball with an initial speed of twice as high as Bill's ball. The maximum height of Joe's ball will be

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Consider the physics of vertical projectile motion under constant gravity. The maximum height h reached by an object launched upward with initial speed u (ignoring air resistance) is h = u^2 / (2g).

Option 1: 'eight times that of Bill's ball.' If Bill's height is h, this would require Joe's height to be 8h, which would correspond to an initial speed u such that u^2/(2g) = 8h. Since h = (Bill’s speed)^2/(2g), Joe would need (8) times the square of Bill’s speed, which is not the case here. This overestimates the height.

Option 2: 'two times that of Bill's ball.' Doubling the height would require Joe’s initial speed to satisfy (u_J)^2 = 2 (u_B)^2, i.e., u_J = sqrt(2) * u_B. But Joe’s speed is stated to be twice Bill’s speed, so this would not match the kinematics.

Option 3: 'equal to that of Bill's ball.' Equal maximum heights would require equal initial speeds, which contradicts the given that Joe’s initial speed is twice Bill’s. Therefore this option cannot be correct.

Option 4: 'four times that of Bill's ball.' Since h ∝ u^2, if Joe’s initial speed is 2 times Bill’s, then h_Joe = (2u_Bill)^2 / (2g) = 4 (u_Bill^2)/(2g) = 4 h_Bill. This matches the squared-dependence of maximum height on initial speed and is consistent with the given information.

In summary, the maximum height scales with the square of the initial speed, so doubling the speed results in quadrupling the height. The only option that reflects this relationship correctly is the one stating four times as high as Bill’s ball.

SPHY001 Practice Quiz - Unit 07 | LA047

Joe and Bill throw identical balls vertically upward. Joe throws his ball with an initial speed of twice as high as Bill's ball. The maximum height of Joe's ball will be

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