An alpine rescue team is using a slingshot to send an emergency medical packet to climbers stranded on a ledge. (height=252 m, horizontal distance=416 m, landing angle= 75° below the horizontal).
What should be the launch speed from the slingshot?
What is the launch angle?
What is the time of flight?
What is the highest point reached by the packet?
you know hf, distance final, df, and the ratio of Vverticalfinal to vhoriz final frm the landing angle. This is going to be an exercise in math.
Lets work it backwards. Assume V as the launching velocity on the ledge, as if they sent a package to you.
hf=hi+ViSin75*time-4.9t^2 or
-252=V Sin 75*t-4.9t^2
in the horizontal
416=Vcos75*t solve for t
t= 416/Vcos75 now put that into the first equation for t, and then solve for V. A bit of algebra is required, get a pad of paper.
Now having V at launch, you have V at the landing (from initial KE+initial PE= final KE)
YOu have the final V, and the horizontal V, so you can find the final angle (theta=arcCos Vhoria/V)
you should be able to take it from here.
for highest point, you have the initial KE, the initial PE, that should equal the KE fromhorizonal+ pE at the top. find max height from that.
To find the launch speed from the slingshot, launch angle, time of flight, and highest point reached by the packet, we can use the equations of projectile motion.
1. Launch speed (initial velocity):
The launch speed can be found using the horizontal distance and the launch angle. The horizontal distance traveled by the packet is 416 m.
Let's calculate the vertical distance traveled by the packet:
Vertical distance (d) = height = 252 m
Since the launch angle is given as 75° below the horizontal, the launch angle relative to the ground (θ) will be 75° + 90° (since the angle is measured below the horizontal) = 165°.
We can now calculate the launch speed using the horizontal and vertical distances:
Horizontal velocity (Vx) = horizontal distance / time of flight
Vertical velocity (Vy) = vertical distance / time of flight
The launch speed (V) can be found using the formula:
V = √(Vx^2 + Vy^2)
2. Launch angle:
The launch angle is given as 75° below the horizontal, which we already converted to 165° relative to the ground.
3. Time of flight:
The time of flight is the total time taken by the packet to reach the ground from the launch point. It can be calculated using the vertical distance and the launch angle, assuming no air resistance.
Time of flight (t) can be found using the formula:
t = 2 * (Vy / g)
where g is the acceleration due to gravity (approximately 9.8 m/s^2).
4. Highest point reached by the packet:
The highest point reached by the packet is the maximum vertical distance from the launch point. It can be calculated using the launch angle and the initial vertical velocity.
Maximum height (H) can be found using the formula:
H = (Vy^2) / (2 * g)
Let's perform the calculations:
1. Launch speed (V):
Vx = 416 m / t (time of flight)
Vy = 252 m / t (time of flight)
V = √(Vx^2 + Vy^2)
2. Launch angle:
Launch angle = 165°
3. Time of flight (t):
t = 2 * (Vy / g)
4. Highest point reached (H):
H = (Vy^2) / (2 * g)
Now, we can substitute the given values and calculate the results.
To find the solution to these questions, we can use the principles of projectile motion. Projectile motion involves the motion of an object that is launched into the air and subject to only the force of gravity. We can break down the motion into the horizontal and vertical components.
1. Launch speed (initial velocity):
To find the launch speed from the slingshot, we can consider the horizontal component of the motion. The horizontal distance traveled by the packet is 416 m, which is the range of motion. We can use the formula for range: R = V₀ * t, where V₀ is the initial horizontal velocity (which is equal to the launch speed) and t is the time of flight. Rearranging the formula, we get V₀ = R / t.
2. Launch angle:
The launch angle is given as 75° below the horizontal. Since we are considering the vertical component of motion, we need to convert this angle to its complement (90° - 75° = 15°) to find the launch angle in relation to the vertical.
3. Time of flight:
To find the time of flight, we can consider the vertical component of the motion. The height of the ledge is given as 252 m. We can use the formula for vertical displacement: Δy = V₀ * sin(θ) * t - 0.5 * g * t², where θ is the launch angle, t is the time of flight, and g is the acceleration due to gravity (approximately 9.8 m/s²). Rearranging the formula, we get t² - (2 * V₀ * sin(θ) / g) * t + (2 * Δy / g) = 0. We can solve this quadratic equation to find the time of flight.
4. Highest point reached:
To find the highest point reached by the packet, we need to determine the maximum vertical displacement, using the same formula as in the previous step. The highest point will occur when the vertical velocity component becomes zero.
By applying these steps, we can calculate the values of the launch speed, launch angle, time of flight, and highest point reached by the packet.