First slide

Projection Under uniform Acceleration

Question

The point from where a ball is projected is taken as the origin of the coordinate axes. The x and y components of its displacement are given by x = 6t and y = 8t - 5 $t^{2}$ . What is the velocity of projection?

Moderate

A

6 ${ms}^{- 1}$
B

8 ${ms}^{- 1}$
C

10 ${ms}^{- 1}$
D

14 ${ms}^{- 1}$

Solution

$v_{x} = \frac{dx}{dt} = 6 and v_{y} = \frac{dy}{dt} = 8 - 10 t$
= 8-10 $\times 0 = 8$
v = $\sqrt{{v_{x}}^{2} + {v_{y}}^{2}} = \sqrt{6^{2} + 8^{2}} = 10 {ms}^{- 1}$

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