A projectile is launched from ground level to the top of a cliff which is 195m away and 135m high. If the projectile lands on top of the cliff 6.9s after it is fired, find the initial velocity of the projectile ((a)magnitude and (b)direction. Neglect air resistance.
The horizontal-component, ux, of the initial velocity is the horizontal distance (195m) divided by 6.9s.
The vertical component, uy, can be found by the equation:
S = uy t + (1/2)(-g)t²
where
S is the vertical distance of the target (135 m)
uy = vertical component of velocity
t = 6.9 seconds
g = acceleration due to gravity (9.81 m/s/s/)
from which uy can be found.
The direction is atan2(uy,ux) from the horizontal.
I don't understand how to get the magnitude. Do you just add the two numbers together?
What do you do to the angle if it asks for the angle below the positive x-axis?
You can obtain the magnitude by doing a vectorial sum, namely magnitude=√(ux²+uy²)
The direction from the x-axis is obtained from tan-1(uy/ux), which goes from -π/2 to +%pi/2. Since the angle could range from 0 to 2π, it is necessary to observe the sign of ux and uy and determine the quadrant accordingly.
For the given question, the angle should be above the horizon (x-axis). If it is below, it will never reach the top of the cliff, and the solution is not correct.
If another questions requires a solution
below the x-axis, the angle will be negative between 0 and π/2.
To find the initial velocity of the projectile, we can break down the motion into vertical and horizontal components.
(a) Finding the horizontal component:
We can use the formula: distance = velocity * time
In the horizontal direction, the distance covered is 195m, and the time taken is 6.9s. Therefore, we can find the horizontal component of the velocity as follows:
horizontal distance = horizontal velocity * time
195m = horizontal velocity * 6.9s
horizontal velocity = 195m / 6.9s
horizontal velocity ≈ 28.26 m/s
(b) Finding the vertical component:
The vertical motion of the projectile can be analyzed using the equations of motion under constant acceleration.
Using the equation: distance = initial velocity * time + (1/2) * acceleration * time^2
For vertical motion, the distance covered is 135m, the initial velocity is unknown (let's call it v0), the time taken is 6.9s, and the acceleration is due to gravity (-9.8 m/s^2).
Plugging in these values, we get:
135m = v0 * 6.9s + (1/2) * (-9.8m/s^2) * (6.9s)^2
Solving this equation will give us the initial vertical velocity, v0.
First, find the (1/2) * (-9.8m/s^2) * (6.9s)^2 term:
(1/2) * (-9.8m/s^2) * (6.9s)^2 = -218.2542
Now, rearrange the equation and solve for v0:
v0 * 6.9s = 135m - 218.2542
v0 * 6.9s = -83.2542
v0 = -83.2542 / 6.9s
v0 ≈ -12.048 m/s
The magnitude of the initial velocity is approximately 12.048 m/s, and since the y-component is negative, the direction is downward (opposite to the positive y-axis).
Therefore, the initial velocity of the projectile is approximately 28.26 m/s in the horizontal direction and approximately 12.048 m/s downward in the vertical direction.