First slide

Projection Under uniform Acceleration

Question

A body is projected at an angle of 30° with the horizontal and with a speed of 30 ms^-1. What is the angle with the horizontal after 1.5 s ? (g = 10 ms^-2)

Easy

A

0°
B

30°
C

60°
D

90°

Solution

The time of flight is given by
$T = \frac{2 usin θ}{g} = \frac{2 \times 30 \times 1 / 2}{10} = 3 s$
Thus, after 1.5 s, the body will be at the highest point.
So the direction of motion will be horizontal after 1.5 s, the angle with the horizontal is 0°.

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