First slide

Projection Under uniform Acceleration

Question

A particle is projected from a horizontal plane with a velocity of $8 \sqrt{2} m / s$ at an angle. At the highest point, its velocity is found to be 8 m/s. Its range will be

Moderate

A

6.4 m
B

3.2 m
C

12.8 m
D

4.6 m

Solution

The velocity at the highest point $= v = v_{0} \cos θ_{0}$
or $8 = 8 \sqrt{2} \cos θ_{0} or θ_{0} = 45^{\circ}$
$\begin{array}{l} Range = \frac{u_{0}^{2} \sin 2 θ_{0}}{g} = \frac{(8 \sqrt{2})^{2} \sin 90^{\circ}}{10} \\ \frac{64 \times 2}{10} = 12 .8 m \end{array}$

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