First slide

Vertical projection from ground

Question

A body is projected vertically upward with a velocity of 30 m/s. It crosses a horizontal line AB at a height above the ground, first during its upward journey and then during its downward journey, in a time interval of 5 sec. Then height of the line AB ‘h’ is equal to

Moderate

A
15 m
B
10.5 m
C
13.75 m
D
20 m

Solution

Time of flight above the line AB is given by T = $\frac{2 v}{g} \Rightarrow v = \frac{g T}{2} = \frac{10 \times 5}{2} m / s = 25 m / s$
Therefore height of line AB h $\frac{u^{2} - v^{2}}{2 g} = \frac{30^{2} - 25^{2}}{2 \times 10} m = 13.75 m$

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