A rock thrown with speed 6.50 m/s and launch angle 30.0 deg(above the horizontal

Question

A rock thrown with speed 6.50 m/s and launch angle 30.0 deg(above the horizontal) travels a horizontal distance of = 12.0 m before hitting the ground. From what height was the rock thrown? Use the value = 9.810 for the free-fall acceleration.

Find the height from which the rock was launched.
Which would be 15.4m

Now a second rock is thrown straight upward with a speed 3.250 m/s . If this rock takes 2.132 s to fall to the ground, from what height H was it released?

Keep in mind that if the positive y axis is upward and the origin is located on the ground, Yi =H.

Express your answer in meters to three significant figures.

Explanation / Answer

I see you already have the first part complete with the answer of 15.4 m. Here is the solution for the second part

Let us find how high it goes above its lauch point

vf2 = vo2 + 2ad

0 = (3.25)2 + (2)(9.81)(d)

d = .538 m

Now, how long does that take.

vf = vo + at

0 = 3.25 + (9.81)(t)

t = .331 sec

Since the total time takes 2.132 sec, then the time to reach the ground from the top of its path is

2.132 - .331 = 1.8 sec

Now we can find how far that distance is

d = vot + .5at2

d = (0) + (.5)(9.81)(1.82)

d = 15.9 m

So, the initial lauch height was 15.9 - .538 = 15.4 m

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