Please answer the following. I\'ll rate, thanks. PSS 4.1 Projectile Motion Probl

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Please answer the following. I'll rate, thanks.

PSS 4.1 Projectile Motion Problems 5 of 10 Express your answer in meters to three significant figures View Available Hint(s) Learning Goal To practice Problem-Solving Strategy 4.1 for projectile motion 6.60 m Previous Answers A rock thrown with speed 12.0 m/s and launch angle 30.0 (above the horizontal) travels a horizontal distance of d 20.0 m before hitting the ground. From what height was the rock thrown? Use the value g 9.800 m/s2 for the free-fall acceleration Correct Significant Figures Feedback: Your answer 6.601 m was either rounded differently or used a different number of significant figures than required for this part. Assess Part D A second rock is thrown straight upward with a speed 6.000 m/s. If this rock takes 1.925 s to fall to the ground, from what height H was it released? Express your answer in meters to three significant figures. View Available Hint(s) Hint 1. Identify the known variables (cick to open) Hint 2. Determine which equation to use to find the height (dick to open) Submit Reguest Answer

Explanation / Answer

2. Strategy: use conservation of energy to find the launch velocity, then basic kinematics.

CoE gives us
initial KE = final KE + final PE
which after dividing through by mass/2 gives
Vi² = Vf² + 2gh
(8m/s)² = Vf² + 2*9.8m/s²*1m
Vf = 6.66 m/s

Now we have the launch velocity, and we know the launch angle = 30º
Vertically, Vy = Vf*sin? = 6.66m/s * sin30º = 3.33 m/s
s = So + Vy*t + ½at²
0 = 1m + 3.33m/s*t - 4.9m/s²*t²
This quadratic has roots at t = -0.22 s ? not possible
and t = 0.905 s ? time of flight

Then horizontally we have
x = Vx*t = Vf*cos?*t = 6.66m/s * cos30º * 0.905s = 5.22 m

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