A daredevil is shot out of a cannon at 45 degrees to the horizontal with an init
ID: 1412953 • Letter: A
Question
A daredevil is shot out of a cannon at 45 degrees to the horizontal with an initial speed of 25 m/s. A net is located a horizontal distance of 50 m from the cannon. a) What is the horizontal component of the daredevil's velocity at its highest point? b) How long does it take to reach its highest point? c) How long is the daredevil in the air? d) At what height above the cannon should the net be placed in order to catch the daredevil? (Someone already answered this, but I think that the answers should be the same number of significant figures)
Explanation / Answer
This involves components and you need to incorporate gravitational acceleration. Here is how I would do it:
Let us break the velocity into the horizontal and vertical components: V_h and V_v. If V is the velocity, then,
V_h = V * Sin 45 = 25/sqrt(2) = 17.6776695 m/s (Google calculator has been used to calculate)
V_v = V * Cos 45 = 25/sqrt(2) = 17.6776695 m/s
The net is at a distance of 50 from the cannon. This is on the same horizontal plane. Let us calculate the time taken to reach at that distance. It will be denoted by T. Since acceleration is zero on the horizontal plane (neglecting air resistance),
T = 50/V_h = 2.82842712 s
Let us see the height of the daredevil at 2.82842712 seconds after launch. Here we will assume that up direction is positive. We will use the following equation:
s = u * t + 0.5 * a * t^2 (s is distance, u is initial velocity, a is acceleration and t is time)
S = V_v * T - 0.5 * g * T^2 = 10.8 m
Thus the net should be at a height of 10.8 m above ground level