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anonymous
 5 years ago
a ball thrown vertically upwards from level ground is observed twice, at a height H above the ground, within a time interval t.The initial velocity of the ball was
a. \[\sqrt{8gH+g ^{2}t ^{2}}\]
b. \[\sqrt{8gH+\left( gT/2 \right)^{2}}\]
c.\[1/2\sqrt{8gH+g ^{2}t ^{2}}\]
d. \[\sqrt{8gH+4g ^{2}t ^{2}}\]
anonymous
 5 years ago
a ball thrown vertically upwards from level ground is observed twice, at a height H above the ground, within a time interval t.The initial velocity of the ball was a. \[\sqrt{8gH+g ^{2}t ^{2}}\] b. \[\sqrt{8gH+\left( gT/2 \right)^{2}}\] c.\[1/2\sqrt{8gH+g ^{2}t ^{2}}\] d. \[\sqrt{8gH+4g ^{2}t ^{2}}\]

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Here is one way to solve it. First we find the maximum height of the ball by the energy principle: \[mgh_{\max}= \frac{1}{2}mv_0^2 \Leftrightarrow h_{\max}= \frac{v_0^2}{2g}\] Then let's denote the distance the ball has dropped from its highest point (after time t/2) with d, and therefore \[d+H=h_{\max}\Leftrightarrow d=h_{max}H\] Because the trajectory is symmetric, we can write \[d=\frac{1}{2}g(\frac{t}{2})^2\] and \[ h_{max}H=\frac{1}{2}g(\frac{t}{2})^2 \] \[ \frac{v_0^2}{2g}H=\frac{1}{2}g(\frac{t}{2})^2 \] from which we can solve \[v_0=\sqrt{\frac{1}{4}(gt)^2+2gH}=\frac{1}{2}\sqrt{(gt)^2+8gh}\] which is answer number c.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0tnx a lot linjaaho !!
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