A projectile is fired at an upward angle of 45 degrees from the top of a 265 m cliff with a speed of 350 m/s. What will be it's speed when it strikes the ground below?
200
To find the speed of the projectile when it strikes the ground, we can use the principles of projectile motion. Firstly, we need to break down the initial velocity of the projectile into its horizontal and vertical components.
The horizontal component of the initial velocity remains constant throughout the motion, and is given by:
Vx = V(initial) * cos(theta)
where V(initial) is the initial velocity (350 m/s) and theta is the launch angle (45 degrees).
Let's calculate Vx:
Vx = 350 m/s * cos(45 degrees)
Vx = 350 m/s * 0.7071
Vx ≈ 247.25 m/s
The vertical component of the initial velocity changes over time due to the effect of gravity. The vertical component of the initial velocity can be calculated as:
Vy = V(initial) * sin(theta)
Let's calculate Vy:
Vy = 350 m/s * sin(45 degrees)
Vy = 350 m/s * 0.7071
Vy ≈ 247.25 m/s
From here, we can determine the time it takes for the projectile to reach the ground using the equation:
h = Vy * t + (1/2) * g * t^2
where h is the height of the cliff (265 m), g is the acceleration due to gravity (approximately 9.8 m/s^2), and t is the time.
Since the projectile is fired from the top of the cliff, its initial vertical position is 0. Therefore, we can simplify the equation to:
0 = Vy * t + (1/2) * g * t^2
Rearranging the equation, we get:
(1/2) * g * t^2 = - Vy * t
Simplifying further, we obtain:
g * t - (2 * Vy) = 0
Solving for t, we find:
t = (2 * Vy) / g
Let's calculate t:
t ≈ (2 * 247.25 m/s) / 9.8 m/s^2
t ≈ 50.49 s
Now, we can use the time of flight to determine the horizontal distance traveled by the projectile:
Range = Vx * t
Let's calculate the range:
Range = 247.25 m/s * 50.49 s
Range ≈ 12,483 m
Since the horizontal distance traveled by the projectile is equal to the range, we can now determine the time it takes for the projectile to hit the ground using the horizontal distance and Vx:
Time = Range / Vx
Let's calculate the time it takes to hit the ground:
Time = 12,483 m / 247.25 m/s
Time ≈ 50.49 s
Since the time to hit the ground is the same as the time of flight, we can now determine the final vertical velocity of the projectile when it reaches the ground:
V(final) = Vy + g * t
Let's calculate V(final):
V(final) = 247.25 m/s + 9.8 m/s^2 * 50.49 s
V(final) ≈ 788.45 m/s
Therefore, the speed of the projectile when it strikes the ground is approximately 788.45 m/s.
the initial energy is 1/2 m v^2 + mgh
the final energy is 1/2 m vf^2 set them equal.
1/2 m vf^2=1/2 m vi^2 + mg265
solve for Vf