Problem 60
Question
A ball is dropped from the top of the Empire State Building. The height, \(y\), of the ball above the ground (in feet) is given as a function of time, \(t\), (in seconds) by $$ y=1250-16 t^{2} $$ (a) Find the velocity of the ball at time \(t\). What is the sign of the velocity? Why is this to be expected? (b) When does the ball hit the ground, and how fast is it going at that time? Give your answer in feet per second and in miles per hour ( \(1 \mathrm{ft} / \mathrm{sec}=15 / 22 \mathrm{mph}\) ).
Step-by-Step Solution
Verified Answer
Velocity at time \( t \) is \(-32t\). The ball hits the ground at \( t \approx 8.84 \) seconds with a speed of \( -282.88 \) ft/s or \(-192.90\) mph.
1Step 1: Understand the problem
We have a ball dropped from a height, and its height above ground is given by the equation \( y = 1250 - 16t^2 \). We need to find the velocity of the ball and the time when it hits the ground.
2Step 2: Differentiate to find velocity
The velocity of the ball, \( v(t) \), is the derivative of the height \( y(t) \) with respect to time \( t \). \[ v(t) = \frac{dy}{dt} = \frac{d}{dt}(1250 - 16t^2) = -32t \].
3Step 3: Determine the sign of velocity
The expression \( v(t) = -32t \) shows that the velocity is negative for all positive \( t \), which indicates that the ball is moving downward (towards the ground). This is expected since the ball was dropped, not thrown upwards.
4Step 4: Solve for ground impact time
To find when the ball hits the ground, set \( y(t) = 0 \): \[ 0 = 1250 - 16t^2 \]. Solve for \( t \): \[ 16t^2 = 1250 \] \[ t^2 = \frac{1250}{16} \] \[ t = \sqrt{\frac{1250}{16}} \approx 8.84 \text{ seconds} \].
5Step 5: Calculate the velocity at impact
Plug \( t = 8.84 \) into the velocity equation: \[ v(8.84) = -32 \times 8.84 \approx -282.88 \text{ ft/s} \]. This negative sign indicates downward motion.
6Step 6: Convert velocity to mph
Convert \(-282.88\) ft/s to mph using the conversion \( 1 \text{ ft/s} = \frac{15}{22} \text{ mph} \): \[ -282.88 \times \frac{15}{22} \approx -192.90 \text{ mph} \].
Key Concepts
Understanding Velocity in PhysicsDifferentiation: Finding Rates of ChangeCalculating Impact Time
Understanding Velocity in Physics
Velocity is a fundamental concept in physics that describes how fast something is moving and in which direction. It is different from speed, as speed does not consider direction. In this exercise, the velocity of the ball is derived from its height equation, representing how the distance above ground changes with time.
The important thing to note here is the sign of the velocity. Because the ball is dropped, it accelerates downwards due to gravity, making the velocity negative, indicating a downward motion. In other contexts, a positive velocity would mean motion in the upwards direction. Thus, the sign of velocity provides valuable information about the direction of movement, and here it confirms the ball's descent.
The important thing to note here is the sign of the velocity. Because the ball is dropped, it accelerates downwards due to gravity, making the velocity negative, indicating a downward motion. In other contexts, a positive velocity would mean motion in the upwards direction. Thus, the sign of velocity provides valuable information about the direction of movement, and here it confirms the ball's descent.
Differentiation: Finding Rates of Change
Differentiation is a powerful tool in calculus that allows us to find the rate at which a quantity changes. In this context, it helps us calculate the ball's velocity at any given time.
To differentiate the given equation for height, \[ y = 1250 - 16t^2 \] we apply the power rule. The derivative, \( v(t) = \frac{dy}{dt}\), is calculated as \[ v(t) = -32t \].
Here, -32 is the constant multiplier that comes from differentiating \[ -16t^2 \] (applying the power rule). This gives us the velocity equation, which shows that the velocity changes linearly with time.
To differentiate the given equation for height, \[ y = 1250 - 16t^2 \] we apply the power rule. The derivative, \( v(t) = \frac{dy}{dt}\), is calculated as \[ v(t) = -32t \].
Here, -32 is the constant multiplier that comes from differentiating \[ -16t^2 \] (applying the power rule). This gives us the velocity equation, which shows that the velocity changes linearly with time.
- If you plug any positive value for \[ t \], you will see that the negative velocity confirms the downward motion.
Calculating Impact Time
When calculating the impact time, we determine when the ball reaches the ground, i.e., when the height becomes zero. To do this, we solve the equation for height: \[ 0 = 1250 - 16t^2 \].
By rearranging and solving, we find that \[ t^2 = \frac{1250}{16} \], so \[ t \approx 8.84 \] seconds, which is the time the ball takes to hit the ground.
Once we have the time, we can find how fast it is going at impact by substituting \[ t = 8.84 \] into the velocity equation \[ v(t) = -32t \], giving us approximately \[ -282.88 \] ft/s. This result can further be converted into miles per hour: \[ -282.88 \times \frac{15}{22} \approx -192.90 \text{ mph} \].
Understanding impact time and how velocity is involved teaches us how objects move under gravity, an essential aspect of physics and calculus.
By rearranging and solving, we find that \[ t^2 = \frac{1250}{16} \], so \[ t \approx 8.84 \] seconds, which is the time the ball takes to hit the ground.
Once we have the time, we can find how fast it is going at impact by substituting \[ t = 8.84 \] into the velocity equation \[ v(t) = -32t \], giving us approximately \[ -282.88 \] ft/s. This result can further be converted into miles per hour: \[ -282.88 \times \frac{15}{22} \approx -192.90 \text{ mph} \].
Understanding impact time and how velocity is involved teaches us how objects move under gravity, an essential aspect of physics and calculus.
Other exercises in this chapter
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