use a(t) = -32 ft per second squared as the acceleration due to gravity.An object is thrown vertically downward from the top of a 480-ft building with an initial velocity of 64 feet per second. With what velocity does the object hit the ground?
This type of problem always shows up on my test and I always get it wrong. Can someone show me how to do it? Thanks.
At this point you should know that
if s(t) is the height obtained, then
v(t) , or velocity, is s'(t) and
a(t) , the acceleration, is v'(t) or s''(t)
Also having done some of these, by now you should know that for the above,
s(t) = -16t^2 + 64t + 480 and
v(t) = -32t + 64
so when it hits the ground, isn't s(t) = 0 ??
so
-16t^2 + 64t + 480 = 0
t^2 - 4t - 30 = 0
t = 2 ± √34 , we can reject the negative value
t = 2+√34
sub that into v(t) and you are done
Certainly! I'd be happy to help explain how to solve this problem. Let's break it down step by step.
First, let's start by finding the time it takes for the object to hit the ground. We can use the equation:
v = u + at
Where:
v = final velocity (which is what we're trying to find)
u = initial velocity (given as 64 ft/s, but since the object is thrown downwards, it will be negative)
a = acceleration due to gravity (given as -32 ft/s^2)
t = time
Since the object is thrown downward, the initial velocity is negative. Therefore, we can substitute the given values into the equation:
0 = (-64) + (-32)t
Now, solve for t:
64 = -32t
t = -64 / -32
t = 2 seconds
Next, we want to find the final velocity (v) when the object hits the ground. We can use the equation:
v = u + at
Using the given values:
v = (-64) + (-32)(2)
v = -64 - 64
v = -128 ft/s
So, the object hits the ground with a velocity of -128 ft/s. Since velocity is a vector quantity, the negative sign indicates that the object is moving downward.
I hope this helps! If you have any more questions, feel free to ask.
To find the velocity with which the object hits the ground, we can use the kinematic equation for vertical motion:
v^2 = u^2 + 2as
where:
- v is the final velocity (the velocity with which the object hits the ground)
- u is the initial velocity (given as 64 ft/sec, but negative since it is downward)
- a is the acceleration due to gravity (given as -32 ft/sec²)
- s is the displacement (the negative of the height of the building, -480 ft)
Let's substitute the given values into the equation:
v^2 = (64 ft/sec)^2 + 2(-32 ft/sec²)(-480 ft)
Before we proceed, let's simplify the equation:
v^2 = 4096 ft²/sec² + 30720 ft²/sec²
v^2 = 34816 ft²/sec²
To find the velocity, we will take the square root of both sides:
v = √34816 ft²/sec²
v ≈ 186.7 ft/sec
So, the object hits the ground with a velocity of approximately 186.7 ft/sec.
Sure! I can help you with that. To find the velocity with which the object hits the ground, we need to use the equations of motion. The equation we will use is:
v^2 = u^2 + 2a s
Where:
- v is the final velocity (which we want to find)
- u is the initial velocity (given as 64 ft/s)
- a is the acceleration due to gravity (-32 ft/s^2)
- s is the displacement or distance traveled (the height of the building, given as 480 ft)
Now, let's plug in the values and solve for v:
v^2 = (64 ft/s)^2 + 2(-32 ft/s^2)(480 ft)
v^2 = 4096 ft^2/s^2 - 30720 ft^2/s^2
v^2 = -26624 ft^2/s^2
Since the velocity must be positive, we take the square root of both sides and get:
v = √(-26624 ft^2/s^2)
Now, here's where it gets tricky. The square root of a negative number does not have a real number solution. However, in physics, we use the imaginary unit "i" to denote the square root of negative numbers. In this case, the velocity will be an imaginary number, indicating that the object never actually hits the ground.
Therefore, with the given initial conditions, the object will never hit the ground.