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Masteriza [31]
3 years ago
15

A student sits atop a platform a distance h above the ground. He throws a pingpong ball horizontally with a speed v. However, a

wind blowing parallel to the ground gives the pingpong ball a constant horizontal acceleration with magnitude a. This results in the pingpong ball reaching the ground directly under the student.
Determine the height h in terms of v, a, and g. This is a somewhat unphysical problem in that you can ignore the effect of air resistance on the vertical motion, but clearly the wind has a big affect on the horizontal motion.
Physics
1 answer:
Nikitich [7]3 years ago
8 0

Answer:

Explanation:

The pingpong ball reaches the ground directly under the student. That means net displacement is zero

Time of fall  t = \sqrt{\frac{2h}{g} }

Initial horizontal velocity = v ,

horizontal acceleration = - a ( deceleration because net displacement is zero)

0 = ut - 1/2 a t²

0 = v\times\sqrt{\frac{2h}{g} } - \frac{1}{2}a \frac{2h}{g}

v = 1/2 a \sqrt{\frac{2h}{g}

h = \frac{2gv^2}{a^2}

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A catapult launches a test rocket vertically upward from a well, giving the rocket an initial speed of 79.6 m/s at ground level.
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Answer:

The rocket is motion above the ground for 44.7 s.

Explanation:

The equations for the height of the rocket are as follows:

y = y0 + v0 · t + 1/2 · a · t²

and, after the engine fails:

y = y0 + v0 · t + 1/2 · g · t²

Where:

y = height of the rocket at time t

y0 =  initial height

v0 = initial speed

t=  time

a = acceleration due to the engines

g = acceleration due to gravity

First, let´s calculate the time the rocket is being accelerated by the engines:

(The center of the framer of reference is located at the ground, y0 = 0).

y = y0 + v0 · t + 1/2 · a · t²

1190 m = 0 m + 79.6 m/s · t + 1/2 · 4.10 m/s² · t²

0 = 2.05  m/s² · t² + 79.6 m/s · t - 1190 m

Solving the quadratic equation:

t = 11.5 s  (the other value of t is discarded because it is negative).

At that time, the engines fail and the rocket starts to fall. The equation of the height will be:

y = y0 + v0 · t + 1/2 · g · t²

The initial velocity (v0) will be the velocity acquired during 11.5 s of acceleration plus the initial velocity of launch:

v = v0 + a · t

v = 79.6 m/s + 4.10 m/s² · 11.5 s

v = 126.8 m/s

Now, we can calculate the time it takes the rocket to reach the ground (y = 0) from a height of 1190 m:

y = y0 + v0 · t + 1/2 · g · t²

0 m = 1190 m + 126.8 m/s · t - 1/2 · 9.80 m/s² · t²

Solving the quadratic equation:

t = 33.2 s

Then, the total time the rocket is in motion is (33.2 s + 11.5 s) 44.7 s

 

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