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Temka [501]
3 years ago
12

A ball is launched directly in the air at a speed of 40 feet per second from a platform located 10 feet in the air. The motion o

f the ball can be modeled using the function f(t)=−16t2+40t+10, where t is the time in seconds and f(t) is the height of the ball. When, in seconds, will the ball hit the ground? Write your answer to the nearest hundredth of a second. Do not include units in your answer.
Physics
1 answer:
crimeas [40]3 years ago
8 0

Answer:

The ball will hit the ground in 2.73 s

Explanation:

Hi there!

f(t) is the height of the ball at time "t". We need to find the value of "t" for which f(t) = 0.

f(t) = 0   Then:

-16 · t² + 40 · t + 10 = 0  

Solving the quadratic equation using the quadratic formula (a = -16, b = 40, c = 10):

t = -0.23 s and t = 2.72 s

Since time can´t be negative, be discard that value.

The ball will hit the ground in 2.73 s.

Have a nice day!

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A bicyclist starting at rest produces a constant angular acceleration of 1.30 rad/s2 for wheels that are 35.5 cm in radius.
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Answer:

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Explanation:

the linear acceleration is given by:

a=\alpha *r\\a=1.30rad/s^2*(35.5*10^{-2}m)\\a=0.462m/s^2

the angular speed is given by:

\omega=\frac{v}{r}\\\\\omega=\frac{11.2m/s}{35.5*10^{-2}m}\\\\\omega=31.5rad/s

to calculate how many radians have the wheel turned we need the apply the following formula:

\theta=\frac{1}{2}\alpha*t^2\\\\t=\frac{\omega}{\alpha}\\\\t=\frac{31.5rad/s}{1.30rad/s^2}\\\\t=24.2s\\\\\theta=\frac{1}{2}*1.30rad/s^2*(24.2s)^2\\\\\theta=381rad

the distance is given by:

d=\theta*r

d=381rad*(35.5*10^{-2}m)\\d=135m

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3 years ago
You have to run 2.2 miles in track. How far is this in feet? Note: There are
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