A daredevil is shot out of a cannon at 45.0° to the horizontal with an initial speed of 28.0 m/s. A net is positioned a horizontal distance of 51.0 m from the cannon. At what height above the cannon should the net be placed in order to catch the daredevil?
9.5
Not gonna tell you the answer directly, but you need to use Trig to find its X & Y velocity. Find the difference between the x axis of the vertex & tramp. Calculate the time according to the horizontal velocity. And take that time multiply it with the gravity
To solve this problem, we need to use the equations of projectile motion.
First, let's break down the motion into horizontal and vertical components. The initial velocity of 28.0 m/s has two components: one in the horizontal direction (Vx) and one in the vertical direction (Vy).
The horizontal component (Vx) remains constant throughout the motion because there is no acceleration in the horizontal direction. Therefore, Vx = 28.0 m/s.
The vertical component (Vy) changes due to the acceleration of gravity (-9.8 m/s^2). Initially, the daredevil is shot upwards, so the initial vertical velocity (Vy) is positive (Vy = 28.0 m/s * sin(45°)). The final vertical velocity (Vf) when the daredevil reaches the net is zero because they come to a stop.
Using the equation:
Vf^2 = Vi^2 + 2 * a * d
Vi = initial velocity
Vf = final velocity
a = acceleration
d = distance
We can solve for d:
0^2 = (28.0 m/s * sin(45°))^2 - 2 * (-9.8 m/s^2) * d
Simplifying the equation:
0 = (28.0 m/s * sin(45°))^2 + 19.6 m/s^2 * d
Substituting the values:
0 = (19.8)^2 + 19.6 * d
Solving for d:
d = - (19.8)^2 / 19.6
d = -19.6 m
Since distance cannot be negative, we ignore the negative value. Therefore, the net should be placed at a height of 19.6 meters above the cannon in order to catch the daredevil.
horizontal:
51=28cos45*t solve for t.
Vertical
h=28sin45*t-4.9t^2
solve for h.