Mechanics - 11

Classical Mechanics Level 2

The gap shown is just for sake of understanding A stone which is at $A$ is projected vertically upwards. It passes through a fixed point $P$ , four seconds after its projection. After three more seconds it again passes through $P$ but this time it is falling towards ground. At what height (in $\text{m}$ ) from the ground is point $P$ located ?

Assume $g = \text{9.8 m/s}^2$

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The answer is 137.2.

1 solution

Steven Chase
Mar 29, 2019

$t_1 = 4 \\ t_2 = 3 \\ g = 9.8 \\ P = v_0 \, t_1 - \frac{1}{2} g \, t_1^2 \\ v_1 = v_0 - g \, t_1 \\ 0 = v_1 \, t_2 - \frac{1}{2} \, g \, t_2^2$

Solve the last equation first for $v_1$ . Then the middle equation for $v_0$ . Then the first equation for $P$ . The value of $P$ comes out to $137.2$

Mechanics - 11

The answer is 137.2.

1 solution

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