A pilot flying low and slow drops a weight; it takes 3.2s to hit the ground, during which it travels a horizontal distance of 155m . Now the pilot does a run at the same height but twice the speed. How much time does it take the weight to hit the ground?
Hint: What affect does horizontal velocity have on vertical fall?
Use: vector displacement equation.
x(t) î + y(t) ĵ = [x(0) + u(0)t] î + [y(0) + v(0)t - ½ gt^2] ĵ
xt î + yt ĵ = (x0 + u0 t) î + (y0 + v0 t - ½ g t2) ĵ
To solve this problem, we can use the concept of time of flight for a projectile. The time of flight, denoted as "T," is the time it takes for an object to reach the ground when dropped from a certain height.
Given that it took 3.2 seconds for the weight to hit the ground when dropped by the pilot flying low and slow, we can use this information to determine the time of flight for the weight.
We know that the horizontal distance traveled by the weight is 155m, and the time of flight is 3.2s. We can use the horizontal distance and time of flight to calculate the initial horizontal velocity (Vx) of the weight using the formula:
Vx = horizontal distance / time of flight
Plugging in the given values, we have:
Vx = 155m / 3.2s
Vx ≈ 48.44 m/s
Now, the question asks what happens when the pilot does a run at the same height but twice the speed. Doubling the speed means that the new horizontal velocity (V'x) will be equal to 2 times the initial horizontal velocity (Vx). Therefore:
V'x = 2 x Vx
V'x = 2 x 48.44 m/s
V'x ≈ 96.88 m/s
Now, we need to find how much time it takes for the weight to hit the ground when dropped with the new horizontal velocity. To do this, we use the formula for time of flight:
T' = horizontal distance / horizontal velocity
Plugging in the known values, we have:
T' = 155m / 96.88 m/s
T' ≈ 1.6 s
Therefore, when the pilot does a run at the same height but twice the speed, it will take approximately 1.6 seconds for the weight to hit the ground.