First slide

Projection Under uniform Acceleration

Question

The height y and the distance x along the horizontal plane of a projectile on a certain planet (with no surrounding atmosphere) are given by y = (8t - 5t²) meter and x = 6t meter, where r is in seconds. The velocity of projection is:

Moderate

A

8 m/s
B

6 m/s
C

10 m/s
D

not obtained from the data

Solution

y = 8t - 5t²
$v_{y} = \frac{dy}{dt} = 8 - 10 t$
x = 6t
$v_{x} = \frac{dy}{dt} = 6$
Velocity at any time ‘t’ $v = \sqrt{v_{x}^{2} + v_{y}^{2}} = \sqrt{6^{2} + {(8 - 10 t)}^{2}}$
Velocity of projection (i.e., velocity at t = 0)
= $\sqrt{6^{2} + 8^{2}} = 10 m / s$

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