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 - 5t². What is the velocity of projection?

Moderate

A

6 ms^-1
B

8 ms^-1
C

10 ms^-1
D

14 ms^-1

Solution

$v_{x} = \frac{dx}{dt} = 6 and v_{y} = \frac{dy}{dt} = 8 - 10 t Putting t = 0 (Since we have to find initial velocity) y_{x} = 8 - 10 \times 0 = 8 v = \sqrt{v_{x}^{2} + v_{y}^{2}} = \sqrt{6^{2} + 8^{2}} = 10 {ms}^{- 1}$

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