A daredevil tries to launch out of a canon and land safely in a safety net. The canon is set at a launch angle of 49.0 degrees to the ground. The intial launch velocity is 27.3 meters/second. How far away should the safety net be placed from the cannon?
find the time it takes to have a final height of zero altitude.
with that time, how far horizontally does it travel?
answer it
u = 27.3 cos 49 forever
Vi = 27.3 sin 49 at start
v = Vi - gt
h = Vi t - (1/2)g t^2
at ground, h = 0
so
t = 2 Vi/g = Vi/4.9
now do it
To find the distance from the cannon to the safety net, we can use the principles of projectile motion.
First, let's break down the initial velocity into its horizontal and vertical components. The horizontal component does not change throughout the motion, while the vertical component is affected by gravity.
The initial vertical velocity (Vy) can be found by multiplying the initial velocity (V) by the sine of the launch angle (θ):
Vy = V * sin(θ)
Plugging in the values gives us:
Vy = 27.3 * sin(49.0°)
Next, we can determine the time it takes for the daredevil to reach the highest point of their trajectory. At the highest point, the vertical velocity will be zero. We can use the following formula to find the time of flight (t):
Vy = g * t
Where g is the acceleration due to gravity, approximately 9.8 m/s². Solving for t:
t = Vy / g
Now, we can find the total time of flight (T) by multiplying t by 2 (since the time it takes to reach the highest point is the same as the time it takes to return to the ground):
T = 2 * t
Next, we find the horizontal distance (X) by multiplying the horizontal component of the initial velocity (Vx) by T:
X = Vx * T
To find Vx, we multiply the initial velocity (V) by the cosine of the launch angle (θ):
Vx = V * cos(θ)
Plugging in the values for V and θ:
Vx = 27.3 * cos(49.0°)
Finally, we can calculate the distance:
X = Vx * T
Plugging in the values for Vx and T, we can find the distance from the cannon to the safety net.