First slide

Projection Under uniform Acceleration

Question

An object is projected so that it just clears two walls of height 7.5 m and with separation 50 m from each other. If the time of passing between the walls is 2.5 s, the range of the projectile will be: (g = 10 m/s²)

Difficult

A

35m
B

70m
C

140 m
D

57.5 m

Solution

$v_{x} = u_{x} = \frac{50}{2 .5} = 20 m / s$
$\frac{2 v_{y}}{g} = 2 .5 \Rightarrow v_{y} = 12 .5 m / s$
Now using, $v_{y}^{2} = u_{y}^{2} - 2 gh$
We have v_y = 17.5 m/s
So, range $= \frac{2 u_{x} v_{y}}{g}$
$\frac{2 (20) (1 .75)}{10} = 70 m$

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