First slide

Projection Under uniform Acceleration

Question

The range of a particle when launched at an angle of $15^{\circ}$ with the horizontal is 1.5 km. What is the range of the projectile when launched at an angle of 45^o to the horizontal

Easy

A

1.5 km
B

3.0 km
C

6.0 km
D

0.75 km

Solution

or $\begin{array}{l} R_{1} = \frac{u^{2} \sin 2 θ}{g} = \frac{u^{2} \sin 30}{g} \\ 1 \cdot 5 = \frac{u^{2}}{2 g} or \frac{u^{2}}{g} = 3 \\ R_{2} = \frac{u^{2} \sin 90}{g} = \frac{u^{2}}{g} = 3 km \end{array}$

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