A stone is thrown from a 50m high cliff and lands 4 seconds later, 40m from the base of the cliff. At what speed and angle was the stone thrown?
The time to fall, T = 4s, tells you the vertical initial velocity Vy0
-50 = Vy0 T - (1/2) gT^2
-50 = Vy0*4 - 4.9* 16 Vy0 - 78.4
4 Vy0 = 28.4
Vy0 = 7.1 m/s (upwards)
The "0" at the end of the subscript means "initial value"
For Vx, solve this equation that uses the horizontal coordinate where it hit the ground:
40 = Vx * T
Vx = 10 m/s (Vx remains the same with time, so it doesn't need a 0 subscript
The speed is sqrt[Vx^2 + Vy0^2]
and the launch angle is arctan Vy0/Vx
To determine the speed and angle at which the stone was thrown, we'll need to break down the motion of the stone into its horizontal and vertical components.
Let's start with the vertical component. We know that the height of the cliff is 50m and the stone took 4 seconds to reach the ground. Using the kinematic equation:
h = ut + (1/2)gt^2
Where h is the height, u is the initial vertical velocity, g is the acceleration due to gravity (approximately 9.8 m/s^2), and t is the time.
Substituting the known values:
50 = u * 4 + (1/2) * (-9.8) * 4^2
Simplifying the equation:
50 = 4u - 78.4
Rearranging to isolate u:
4u = 50 + 78.4
4u = 128.4
u = 128.4 / 4
u = 32.1 m/s
So, the initial vertical velocity (upwards) of the stone was 32.1 m/s.
Now let's calculate the horizontal component. We know that the horizontal displacement is 40m, and the time taken is 4 seconds. The horizontal distance traveled can be found using the equation:
s = ut
Where s is the horizontal distance, u is the horizontal initial velocity, and t is the time.
Substituting the known values:
40 = u * 4
Simplifying the equation:
u = 40 / 4
u = 10 m/s
So, the initial horizontal velocity of the stone was 10 m/s (constant since there is no horizontal acceleration).
To find the total initial velocity of the stone, we can use the Pythagorean theorem:
v = √(u^2 + v^2)
Substituting the values:
v = √(10^2 + 32.1^2)
v = √(100 + 1030.41)
v = √1130.41
v ≈ 33.63 m/s
Therefore, the initial speed at which the stone was thrown is approximately 33.63 m/s.
Finally, we can calculate the angle at which the stone was thrown using trigonometry. The angle can be determined using the inverse tangent function:
θ = tan^(-1) (u/v)
Substituting the values:
θ = tan^(-1) (32.1/10)
θ ≈ 73.74°
So, the stone was thrown at an angle of approximately 73.74° to the horizontal.