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A bowling ball is dropped from a height of 50 feet

a. Write a function that gives the height h (in feet) of the bowling ball after t seconds.

h(t)=

b. Find the height of the bowling ball after 1 second.



c. How long does the bowling ball take to hit the ground? Round your answer to the nearest hundredth of a second.

User Sergiom
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1 Answer

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Answer:

A. The function that gives the height of the bowling ball after t seconds is
h(t) = 50-16.087\cdot t^(2)\,\,\,[ft].

B. The height of the bowling ball after 1 second is 33.913 feet.

C. The bowling ball will take approximately 1.763 seconds to hit the ground.

Explanation:

A. From Physics we notice that ball experiments a free-fall, which is an uniform accelerated motion due to gravity and where air friction and effects of the Earth's rotation are neglected. The height of the ball is described by the following kinematic equation:


h(t) = h_(o)+v_(o)\cdot t +(1)/(2)\cdot g \cdot t^(2) (1)

Where:


h_(o) - Initial height of the ball, measured in feet.


v_(o) - Initial velocity of the ball, measured in feet per second.


t - Time, measured in seconds.


g - Gravitational acceleration, measured in feet per square second.

If we know that
h_(o) = 50\,ft,
v_(o) = 0\,(ft)/(s) and
g = -32.174\,(ft)/(s^(2)), then the function that gives the height of the bowling ball after t seconds is:


h(t) = 50-16.087\cdot t^(2)\,\,\,[ft] (2)

The function that gives the height of the bowling ball after t seconds is
h(t) = 50-16.087\cdot t^(2)\,\,\,[ft].

B. If we know that
t = 1\,s, then the height of the bowling ball is:


h(1) = 50-16.087\cdot (1)^(2)


h(1) = 33.913\,ft

The height of the bowling ball after 1 second is 33.913 feet.

C. The bowling ball hits the ground when height is zero. Let equalize (1) to zero and solve the resulting expression:


50-16.087\cdot t^(2) = 0


16.087\cdot t^(2) = 50


t \approx 1.763\,s

The bowling ball will take approximately 1.763 seconds to hit the ground.

User SirRichie
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