A daredevil is shot out of a cannon at 36.5◦ to the horizontal with an initial speed of 36.1 m/s. A net is positioned a horizontal distance of 25.7 m from the cannon. At whatheight abovethecannon shouldthe net be placed in order to catch the daredevil? The acceleration of gravity is 9.8 m/s2 . Answer in units of m
To find the height above the cannon where the net should be placed to catch the daredevil, we can use the principle of projectile motion. The daredevil is launched at an angle of 36.5° with an initial speed of 36.1 m/s. We need to find the vertical distance (height) at which the net should be placed.
Let's break the motion of the daredevil into horizontal and vertical components:
Horizontal Component:
The initial horizontal velocity is given by:
Vx = V * cos(θ) = 36.1 m/s * cos(36.5°)
Vertical Component:
The initial vertical velocity is given by:
Vy = V * sin(θ) = 36.1 m/s * sin(36.5°)
Using the vertical motion equation:
y = yo + Vy * t - 0.5 * g * t^2
We know that the daredevil will pass a horizontal distance of 25.7 m, which is the same as the time of flight multiplied by the horizontal velocity:
t = (distance) / (horizontal velocity) = 25.7 m / Vx
Substituting this value of t back into the equation for y, we can solve for the height above the cannon where the net should be placed (y):
y = Vy * t - 0.5 * g * t^2
Now we can calculate the values and find the answer:
Vx = 36.1 m/s * cos(36.5°)
Vy = 36.1 m/s * sin(36.5°)
t = 25.7 m / Vx
y = Vy * t - 0.5 * g * t^2
Plugging in the values and solving the equation will give us the answer in units of meters.