What Are the Velocities and Displacement of a Projectile Launched from a Cliff?

[ken62310]

A projectile is launched from the edge of a 100m high vertical cliff at an angle of 30 degree below the horizontal. The projectile hits the ground at a distance of 50m from the bottom of the cliff.
-Find the initial and final velocities of the projectile(magnitudes and directions).
-Find total displacement performed by the projectile.

 

[Kurdt]

Welcome to PF. What have you attempted so far? I assume you are familiar with projectile equations.

 

[m2003]

To solve this question, we can use the principles of projectile motion and basic kinematics equations. First, we need to break down the given information into known variables. The initial height of the projectile (H) is 100m, the horizontal distance traveled (x) is 50m, the launch angle (θ) is 30 degrees, and the acceleration due to gravity (g) is 9.8m/s^2.

To find the initial velocity (Vi), we can use the equation Vi^2 = Vx^2 + Vy^2, where Vx is the initial horizontal velocity and Vy is the initial vertical velocity. We can find Vx using the equation Vx = Vi cosθ, and since the projectile is launched at an angle below the horizontal, we can substitute cosθ with sin(90-θ). Therefore, Vx = Vi sin60 = 0.87Vi. To find Vy, we can use the equation Vy = Vi sinθ, which becomes Vy = Vi sin30 = 0.5Vi. Substituting these values into the initial velocity equation, we get Vi = √(Vx^2 + Vy^2) = √(0.87Vi)^2 + (0.5Vi)^2) = 1.03Vi. Therefore, the initial velocity of the projectile is 1.03 times the unknown value of Vi.

To find the final velocity (Vf), we can use the equation Vf^2 = Vx^2 + Vy^2, where Vx is the final horizontal velocity and Vy is the final vertical velocity. Since the projectile hits the ground, Vy = -Vi, and Vx remains the same as the initial horizontal velocity. Therefore, Vf^2 = (0.87Vi)^2 + (-0.5Vi)^2 = 0.76Vi^2. Taking the square root of both sides, we get Vf = 0.87Vi. The final velocity is 0.87 times the initial velocity.

To find the total displacement of the projectile, we can use the equation x = Vx t, where t is the time taken for the projectile to hit the ground. We can find t using the equation H = Vy t + 0.5gt^2, where H is the initial height. Substituting the known values, we get 100 = (-0.5Vi) t