First slide

Projection Under uniform Acceleration

Question

The range of a proiectile, when launched at an angle of $15^{0}$ with the horizontal is 1.5 km. What is the range of the projectile when launched at an angle of 45° to the horizontal ?

Easy

A

1.5 km
B

3 km
C

6 km
D

0.75 km

Solution

$H_{m a x} = \frac{u^{2} \sin^{2} θ}{2 g} \Rightarrow 1 .5 \times 10^{3} = \frac{u^{2} \sin 30 °}{g} = \frac{u^{2}}{g} = 3 \times 10^{3}$
$At θ = 45 °, R = \frac{u^{2}}{g} = 3 \times 10^{3} m = 3 km$

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