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NeTakaya
3 years ago
9

Two identical balls are thrown vertically upward. The second ball is thrown with an initial speed that is twice that of the firs

t ball. How does the maximum height of the two balls compare?
Physics
1 answer:
9966 [12]3 years ago
8 0

Answer:

The maximum height of ball 2 is 4 times that of ball 1

Explanation:

We can find the maximum height of each ball by using the following suvat equation:

v^2-u^2=2as

where

v is the final velocity

u is the initial velocity

a=g=-9.8 m/s^2 is the acceleration of gravity (we take upward as positive direction)

s is the displacement

At the maximum height, s = h and v = 0 (the final velocity is zero), so re-arranging the equation:

h=\frac{-u^2}{2g}

The first ball is thrown with initial velocity u_1, so it reaches a maximum height of

h_1 = -\frac{u_1^2}{2g} (the quantity will be positive, since g is negative)

The second ball is thrown with initial velocity

u_2 = u_1

so it will reach a maximum height of

h_2 = - \frac{u_2^2}{2g}=-\frac{(2u_1)^2}{2g}=4(-\frac{u_1^2}{2g}) = 4h_1

So, its maximum height will be 4 times the maximum height reached by ball 1.

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Negative 9.8 meters per second squared

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3 years ago
Sam receives the kicked football on the 3 yd line and runs straight ahead toward the goal line before cutting to the right at th
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Answer:

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Explanation:

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3 years ago
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