A building is 50 meters tall. If you drop a penny from the roof, its height after t seconds is given s(t)=100-4.9t2 in meters (until it hits the ground)
a) Compute the velocity of the penny after t seconds (before it hits the ground)
b) Compute the acceleration of the penny after t seconds (before it hits the ground).
c) At which time does the penny hit the ground?
d) What is the velocity of the penny when it hits the ground?
The Correct Answer and Explanation is :
Let’s go through each part of the problem step by step.
Given Information:
The height of the penny after ( t ) seconds is given by the equation:
[
s(t) = 100 – 4.9t^2
]
This is a quadratic equation that describes the motion of an object under the influence of gravity. The equation tells us that at ( t = 0 ), the penny starts from a height of 100 meters, and its height decreases as time progresses due to the acceleration of gravity.
Part a) Compute the velocity of the penny after ( t ) seconds.
Velocity is the rate of change of the object’s position with respect to time. To find the velocity function ( v(t) ), we need to take the derivative of the height function ( s(t) ) with respect to time:
[
v(t) = \frac{ds(t)}{dt}
]
Let’s compute this derivative:
[
v(t) = \frac{d}{dt}(100 – 4.9t^2)
]
The derivative of 100 is 0, and the derivative of ( -4.9t^2 ) is ( -9.8t ). Thus:
[
v(t) = -9.8t
]
This equation tells us that the velocity of the penny is proportional to the time, with the negative sign indicating that the penny is falling downward.
Part b) Compute the acceleration of the penny after ( t ) seconds.
Acceleration is the rate of change of velocity with respect to time. To find the acceleration ( a(t) ), we take the derivative of the velocity function ( v(t) ) with respect to time:
[
a(t) = \frac{dv(t)}{dt}
]
Since ( v(t) = -9.8t ), its derivative is:
[
a(t) = -9.8
]
This shows that the acceleration of the penny is constant at ( -9.8 \, \text{m/s}^2 ), which is the acceleration due to gravity near the Earth’s surface. The negative sign indicates that the acceleration is downward.
Part c) At which time does the penny hit the ground?
The penny hits the ground when its height ( s(t) ) reaches 0 meters. So, we set the height function equal to 0 and solve for ( t ):
[
s(t) = 100 – 4.9t^2 = 0
]
Solving for ( t ):
[
4.9t^2 = 100
]
[
t^2 = \frac{100}{4.9}
]
[
t^2 \approx 20.41
]
[
t \approx \sqrt{20.41} \approx 4.51 \, \text{seconds}
]
Thus, the penny hits the ground at approximately ( t \approx 4.51 ) seconds.
Part d) What is the velocity of the penny when it hits the ground?
To find the velocity when the penny hits the ground, we substitute ( t = 4.51 ) seconds into the velocity function ( v(t) = -9.8t ):
[
v(4.51) = -9.8 \times 4.51 \approx -44.95 \, \text{m/s}
]
Thus, the velocity of the penny when it hits the ground is approximately ( -44.95 \, \text{m/s} ), indicating that the penny is falling downward at a speed of about 44.95 meters per second.
Summary of Answers:
- a) The velocity of the penny after ( t ) seconds is ( v(t) = -9.8t ).
- b) The acceleration of the penny is constant and equal to ( a(t) = -9.8 \, \text{m/s}^2 ).
- c) The penny hits the ground at approximately ( t = 4.51 ) seconds.
- d) The velocity of the penny when it hits the ground is approximately ( v \approx -44.95 \, \text{m/s} ).
This problem demonstrates the basic principles of motion under gravity, with the penny falling from a height and undergoing constant acceleration due to Earth’s gravitational pull.