Relative velocity and projectile motion problem solving

SAM31 · Sep 19, 2022

Handwritten physics solution part a: displacement Δd_y = 4.3 m [down], acceleration a = 9.8 m/s² [down], initial velocity v_i = 30.0 m/s [up]; then v_ix = v_i cos θ = (30.0 m/s)(cos 35°) = 24.6 m/s [right]; v_iy = v_i sin θ = (30.0 m/s)(sin 35°).

Handwritten-style kinematics working on a projectile problem: vertical velocity values, a displacement equation solved with the quadratic formula to give t = 3.7 s, and a horizontal range calculation beginning Δd_x = v_x Δt.

Handwritten-style worked solution of a projectile motion problem showing range 92 m right, horizontal velocity 24.6 m/s right, vertical velocity 19 m/s down, and final speed 31 m/s.

Handwritten-style typed working: tan θ = |v_fy| / |v_fx|, then tan θ = 19.48937325 m/s / 24.57456133 m/s, θ = tan⁻¹ of that ratio, θ = 38.41691594°, θ = 38°, followed by the line: the ball is moving towards the right and downwards, and the velocity is 31 m/s [right 38° down].

kuruman · Sep 19, 2022

It seems that you have a complete solution. Is there a question you wish to ask or are you posting a solution for the benefit of others?

SAM31 · Sep 19, 2022

kuruman said:

It seems that you have a complete solution. Is there a question you wish to ask?

I was wondering if someone could look over the steps to ensure I'm on the right track and provide me with any feedback!

kuruman · Sep 19, 2022

Is this your solution or someone else's? If yours, it looks professional. I have not put in the numbers, but it looks right.

SAM31 · Sep 19, 2022

kuruman said:

Is this your solution or someone else's?

This is my own attempt at a solution for the question asked above

kuruman · Sep 19, 2022

See edited post #4. If you wish to check your work, plug in the time of flight in the kinematic equations and see if the ball lands where you thought it would be.

pbuk · Sep 20, 2022

Arithmetic errors creep in here:

Handwritten-style working of the quadratic formula solving for time, starting from Δt = (-17.20729309 m/s ± √((17.20729309 m/s)² - 4(-4.9 m/s²)(4.3 m))) / (2(-4.9 m/s²)), then Δt = (-17.20729309 ± √101.4872931) / -9.8, giving Δt = 3.74455779 s or Δt = -0.23453371 s, and finally Δt = 3.7 s.

but they do not affect the final results to 2sf, and otherwise it looks good.

PeroK · Sep 20, 2022

SAM31 said:

This is my own attempt at a solution for the question asked above

All those long decimal numbers don't look good to me. This is why algebra was invented.

Relative velocity and projectile motion problem solving

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