Question

Now we are going to look at a classic example of projectile motion: a bullet fired from a gun. A paintball is fired horizontally at a speed of 75.0 m/s from a point 1.50 m above the ground. The ball misses its target and hits the ground some distance away. (a) For how many seconds is the ball in the air? (b) Find the maximum horizontal displacement (which we’ll call the range of the ball). Ignore air resistance.

Answer 1

Answer:

(a) 0.553 s

(b) 41.5 m

Explanation:

(a) We consider the motion in the vertical and horizontal directions. Both occur in the same time when the ball hits the ground. Since the initial speed is fully horizontal, the vertical motion has no initial speed and is simply as the ball was dropped. We use the equation of motion:

$s = ut+\frac{1}{2} at^2$

$s$ is the distance travelled, in this case, the height

$u$ is initial speed

$t$ is the time

$a$ is the acceleration, in this case, acceleration of gravity, $g$

$1.5 \text{ m} = (0\text{ m/s}) \times t + \frac{1}{2}(9.81 \text{ m/s}{}^2)\times t^2\\t = \sqrt{\dfrac{3 \text{ m}}{9.8 \text{ m/s}{}^2}} = 0.553 \text{ s}$

(b) The maximum horizontal displacement is determined by the horizontal motion which is non-accelerated. This displacement is the product of the horizontal speed and time (from (a) above).

$d = 75\text{ m/s}\times0.553 \text{ s} = 41.5 \text{ m}$