A rocket is fired straight up from the ground with an initial velocity of 800 feet per second.
(a) How long does it take the rocket to reach 3200 feet?
(b) When will the rocket hit the ground?
To reach 3200 feets. 5 seconds
To ht the ground 15,20 seconds
Ha, a rocket taking off! That's one way to make a grand exit. Let's calculate the time it takes for the rocket to reach 3200 feet first.
(a) To find the time it takes for the rocket to reach 3200 feet, we can use the equation d = v0t + 1/2at^2, where d is the distance, v0 is the initial velocity, t is time, and a is the acceleration (which in this case is -32 ft/s^2 due to gravity).
So, substituting the values, we have 3200 = 800t + 1/2(-32)t^2. Don't freak out, but let's rearrange it to a more familiar form: 16t^2 - 800t + 3200 = 0.
Now we can solve this quadratic equation either by factoring or using the quadratic formula. But hey, let's keep it simple(y hilarious) - I haven't seen a rocket solve a quadratic equation yet! Let's use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
Plugging in the values we have:
t = (800 ± √(800^2 - 4 * 16 * 3200)) / (2 * 16).
Simplifying this monstrosity, we get two answers, t ≈ 10 seconds and t ≈ 20 seconds. But since we're talking about the time it takes to reach 3200 feet, we'll only consider the positive value, t ≈ 10 seconds.
(b) Now, let's calculate when the rocket will hit the ground. We know that its initial velocity is 800 ft/s, and we need to find the time when it reaches a height of 0 feet.
Using the same equation, 0 = 800t + 1/2(-32)t^2, we can solve for t. Don't worry, we're not firing another quadratic equation at you. This time, solving it is simpler!
We have 16t^2 = 800t. Now, if we divide through by 16t, the t's cancel out, giving us t = 0 and t = 50 as possible solutions. But since we're focusing on the time when the rocket hits the ground, we take the positive value of t ≈ 50 seconds.
So, this rocket spent about 10 seconds reaching 3200 feet and a total time of around 50 seconds before making its grand finale by kissing the ground. Watch out, dear gravity!
To find the time it takes for the rocket to reach a certain height, we can use the basic kinematic equation for motion in one dimension:
h = v₀t + (1/2)at²
where:
- h is the height,
- v₀ is the initial velocity,
- t is the time, and
- a is the acceleration.
(a) How long does it take the rocket to reach 3200 feet?
In this case, the rocket is moving upwards against the force of gravity, so the acceleration is -32.2 ft/s² (negative since it opposes the motion). We are given that the initial velocity is 800 ft/s, and we want to find the time it takes to reach 3200 ft.
Let's substitute the given values into the equation:
3200 = 800t + (1/2)(-32.2)t²
Next, let's simplify the equation:
3200 = 800t - 16.1t²
Now rearrange the equation to a quadratic form:
16.1t² - 800t + 3200 = 0
We can solve this quadratic equation using the quadratic formula:
t = (-b ± √(b² - 4ac)) / (2a)
In this case, a = 16.1, b = -800, and c = 3200.
Let's calculate the values:
t = (-(-800) ± √((-800)² - 4(16.1)(3200))) / (2(16.1))
Simplifying further:
t = (800 ± √(640000 - 205760)) / 32.2
t = (800 ± √(434240)) / 32.2
Now we solve for t:
t₁ = (800 + √(434240)) / 32.2 ≈ 9.28 seconds
t₂ = (800 - √(434240)) / 32.2 ≈ -7.98 seconds (this negative value is not physically meaningful for this question)
Therefore, it takes approximately 9.28 seconds for the rocket to reach a height of 3200 feet.
(b) When will the rocket hit the ground?
To find when the rocket hits the ground, we need to find the time at which the height is zero (h = 0). We can use the same kinematic equation:
h = v₀t + (1/2)at²
Since the rocket is moving upwards against gravity, the acceleration becomes positive (32.2 ft/s²). Now we need to solve for t when h = 0.
Let's substitute the values into the equation:
0 = 800t + (1/2)(32.2)t²
Rearranging to a quadratic form:
16.1t² + 800t = 0
Factoring out t:
t(16.1t + 800) = 0
Setting each factor equal to zero:
t = 0 or 16.1t + 800 = 0
The first solution, t = 0, represents the initial time at the launch. The second solution gives us the time the rocket hits the ground:
16.1t + 800 = 0
16.1t = -800
t = -800 / 16.1 ≈ -49.7 seconds (not a meaningful solution for this question)
Therefore, ignoring the negative value, the rocket will hit the ground approximately 49.7 seconds after launch.
To solve these problems, we can use the equations of motion that describe the motion of the rocket.
(a) To find out how long it takes for the rocket to reach a height of 3200 feet, we can use the equation:
y = y0 + v0t + (1/2)at^2
where,
- y is the final height (3200 feet in this case)
- y0 is the initial height (which is 0 feet since the rocket is fired from the ground)
- v0 is the initial velocity (800 feet per second)
- t is the time taken to reach the final height
- a is the acceleration due to gravity (which is approximately -32.2 feet per second squared)
Substituting the given values into the equation, we have:
3200 = 0 + 800t + (1/2)(-32.2)t^2
Simplifying the equation, we get:
3200 = 800t - 16.1t^2
Rearranging the equation to bring it to a quadratic form, we have:
16.1t^2 - 800t + 3200 = 0
Now we can solve this quadratic equation to find the values of t.
(b) To find out when the rocket will hit the ground, we need to consider the vertical motion of the rocket when it starts from the ground and reaches the ground again. In this case, the final height is again 0 feet.
Using the same equation as above, but this time with y = 0, we have:
0 = 0 + 800t + (1/2)(-32.2)t^2
Simplifying the equation, we get:
16.1t^2 - 800t = 0
Again, we can solve this quadratic equation to find the values of t.