Question

A snowball is thrown from ground level (by someone in a hole) with initial speed${\mathit{v}}_{\mathbf{0}}$ at an angle of$\mathbf{45}\mathbf{°}$ relative to the (level) ground, on which the snowball later lands. If the launch angle is increased, do (a) the range and (b) the flight time increase, decrease, or stay the same?

Short answer

1. Range of snow ball decreases as the angle increases.
2. Time of flight of snow ball increases as the angle increases.

Step-by-step solution

Given information

Here angle is$\theta =45°$and initial is$v_0$ .

To understand the concept

This problem is based on the concept of projectile motion. When a particle is thrown near the earth’s surface, it travels along a curved path under constant acceleration directed towards the center of the earth surface. Using concept of projectile, the range of projectile can be found. For range, vertical and horizontal components of initial velocity as well as vertical and horizontal distances can be used. Also using the kinematic equations the time of flight of projectile can be found.

Formulae:

$R=\frac{v_0^2\times \sin \left(2\theta \right)}{g}$

$t=\frac{2v_{0}s\mathrm{in}\left(\theta \right)}{g}$

(a) To find the range of snow ball

At$\theta =45°$ every projectile has maximum range

We can find the range R of the projectile by plugging the values of initial velocity and angle.

As the angle increased the range will decrease because of$\sin \left(2\theta \right)$ . As angle increases, the value of sin$\left(2\theta \right)$ will decrease.

(b) To find the time of flight of snow ball

If the angle is increased the value of t increases as the higher angles value of sin tends to 1.