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Mazyrski [523]
3 years ago
7

If a basball is project upwards from the ground level with an initial velovaity of 32 feet per second, then it's height is a fun

ction of time.
Given by s= -8t^2 + 32t. what is the maximum height reached by the ball?
Physics
1 answer:
inessss [21]3 years ago
6 0

Answer:

Maximum height reached by the ball is 32 meters.

Explanation:

It is given that,

If a baseball is project upwards from the ground level with an initial velocity of 32 feet per second, then it's height is a function of time. The equation is given as :

s=-8t^2+32t...........(1)

t is the time taken

s is the height attained as a function of time.

Maximum height achieved can be calculated as :

\dfrac{ds}{dt}=0

\dfrac{d(-8t^2+32t)}{dt}=0

-16 t + 32 = 0

t = 2 seconds

Put the value of t in equation (1) as :

s=-8(2)^2+32(2)

s = 32 meters

So, the maximum height reached by the ball is 32 meters. Hence, this is the required solution.

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A rocket ship has several engines and thrusters. While the Solid Rocket Booster (SRB) and main engines only work together during
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A rocket ship is accelerated by the SRB and the main engines for 2.0 minutes and the main engines for 8.5 minutes after the launch. The acceleration of the ship during the first 2.0 minutes is 11 m/s² (D).

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We want to know the acceleration in the first part (first 2.0 minutes). We need to consider that:

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The acceleration of the ship during the first 2.0 minutes is:

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A rocket ship is accelerated by the SRB and the main engines for 2.0 minutes and the main engines for 8.5 minutes after the launch. The acceleration of the ship during the first 2.0 minutes is 11 m/s² (D).

Learn more: brainly.com/question/16274121

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The gravitational potential energy of an object is equal to its weight multiplied by its
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Prove(show) ''T=2π√(l/g)''​
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T=2\pi\sqrt{\frac{l}{g}}

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