Answer:
A bullet is fired horizontally from a rifle 1.5m from the ground at 430m/s. How far does it travel and for how long does it travel before it hits the ground?
Assuming that air resistance is negligable
What's important to realize about this question is that the horizontal velocity of the bullet makes no difference to how long it takes to fall to the the ground.
Because the gun has not applied any vertical forces to the bullet, the only force affecting the bullet is gravity. This means the bullet takes just as long to fall to the ground as it would if it were dropped, despite it now travelling a large horizontal distance in the duration.
To find the travel time before hitting the ground we have 3 values:
-The displacement from the ground that the bullet must travel, s = 1.5m
-The acceleration the bullet experiences. As gravity is accelerating the bullet downwards, a = g = ~9.81m/s^2
-The initial velocity of the bullet vertically. As the bullet is stationary vertically (it is only travelling horizontally at the start), u = 0m
We examine our equations of motion, commonly known as SUVAT equations. You may need to learn these for your exam, but some examination boards provide them.
Because we have s, u and a, and we are looking for the time t, the relevant equation is
s = ut + 0.5(at^2)
Filling in our values we have:
1.5 = 0t + 0.5(9.81 x t^2)
1.5 = 4.905 x t^2
Divide 1.5 by 4.905 to find t^2
t^2 = 0.3058...
We simply find the square root of t^2 to find t, the time taken for the bullet to reach the ground:
t = 0.553s (3 significant figures)
To find the horizontal distance, d, that the bullet has travelled before it has hit the ground we can use the equation linking displacement s with some velocity v over some time duration t:
s = vt
The horizontal velocity of the bullet, v = 430
The time before the bullet hits the ground, t = 0.553
So d = vt = 430 * 0.553 = 238m (3 significant figures)
Explanation:
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