First slide

Projection Under uniform Acceleration

Question

A projectile is thrown at a angle of 15^o with the horizontal and the range is 1.5 km. What is the range when it is projected at 45^o ?

Easy

A

1.5 km
B

6 km
C

4.5 km
D

3 km

Solution

The range R of the projectile is given by
$R = \frac{u^{2} \sin 2 θ}{g}$
In first case, $1 \cdot 5 = \frac{u^{2} \sin (2 \times 15)}{g} = \frac{u^{2}}{2 g}$
In second case, $R = \frac{u^{2} \sin (2 \times 45)}{g} = \frac{u^{2}}{g}$
$= 1 \cdot 5 \times 2 = 3 km$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App