Question

A bullet is fired into the air at an angle of 30° to the horizontal. At the same time and from the same height,

a bullet is dropped. If we neglect the effects of air resistance, how will the time the two bullets spend in the

air compare?

Answer 1

Answer:

The bullet that is fired will spend longer  in the air, hitting the ground after the  dropped bullet.

Explanation:

Using the equation: x =  x 0  +   v t

If we neglect the effects of air resistance, the horizontal motion is a constant velocity.

The horizontal displacement = (velocity X cosθ)

So, the fired bullet has to travel horizontally before falling which takes a  longer time compared to a bullet dropped where it is, height = 1/2 gt^2

gravity, g = 9.8 m/s2.

Answer 2

The time spent in air for the bullet projected at angle depends on the initial velocity of the bullet and angle of projection while bullet dropped from the same height depends on the final velocity of the bullet which depends on the height from the which the bullet was

The given parameters;

• angle of the projection, θ = 30⁰

The time taken for the bullet fired at angle to return to the ground level is calculated as;

$T = \frac{2 \times usin(30)}{g} \\\\T = \frac{u}{g}$

where;

• u is the initial velocity of the bullet

The time taken for bullet dropped from the same height to reach the ground is calculated as;

$v_f = v_0 + gt\\\\t = \frac{v_f - v_0}{g} \\\\t = \frac{v_f}{g}$

where

• $v_f$ is the final velocity of the bullet

Thus, we can conclude that the time spent in air for the bullet projected at angle depends on the initial velocity of the bullet and angle of projection while bullet dropped from the same height depends on the final velocity of the bullet which depends on the height from the which the bullet was dropped.

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