Question

Based on the data collected, explain why a launch angle of 30 degrees (HIGH HORIZONTAL VELOCITY & SMALL TIME OF FLIGHT) will allow a projectile to go just as far when the launch angle of 60 degrees (LOW HORIZONTAL VELOCITY AND LARGE TIME OF FLIGHT).

Answer 1

Answer:

Explanation:

The horizontal distance traveled by the projectile is given by the formula

$R=\frac{u^{2}Sin2\theta }{g}$

The formula for the time of flight is given by

$T = \frac{2u Sin\theta }{g}$

Case I: when the launch angle is 30°

So, $R_{1}=\frac{u^{2}Sin60 }{g}$

$R_{1}=\frac{0.866u^{2}}{g}$

Horizontal velocity = u Cos 30 = 0.866 u

$T_{1} = \frac{2u Sin30 }{g}=\frac{u}{g}$

Case II: when the launch angle is 60°

$R_{2}=\frac{u^{2}Sin120}{g}$

$R_{2}=\frac{0.866u^{2}}{g}$

Horizontal velocity = u Cos 60 = 0.5 u

$T_{1} = \frac{2u Sin60 }{g}=\frac{1.73 u}{g}$

By observing the case I and case II, we conclude that

R1 = R2

Horizontal velocity 1 > Horizontal velocity 2

T1 < T2