The height in metres of a projectile launched from the top of a building is given by h(t) =- 5t2 + 10t+5 where t is the time in seconds since it was launched. a) How high was the projectile at the moment of launch? b) At what time does the projectile hit the ground? c) What is the average rate of change in height from the time the object was launched until the time it hit the ground? d) What is the velocity of the projectile when it's at its maximum height. Justify your answer algebraically. [5A]
a) when t = 0, h = 5 meters
b) when is h = 0 (besides before the launch :)
0 = - 5 t^2 + 10 t+ 5
solve quadratic
t = -.41 and +2.41 seconds
use 2.41
c)change in height = - 5 meters
change in time =2.41 s
so
v av = -5/2.41
d)zero of course, it is stopped at the top (unless it also had horizontal speed)
I suppose you could find the vertex of the parabola
a) At the moment of launch, t = 0. Substituting t = 0 into the equation h(t) = -5t^2 + 10t + 5, we get:
h(0) = -5(0)^2 + 10(0) + 5
h(0) = 0 + 0 + 5
h(0) = 5
Therefore, the projectile was 5 metres high at the moment of launch.
b) To find the time when the projectile hits the ground, we set h(t) = 0 and solve for t:
0 = -5t^2 + 10t + 5
Rearranging:
5t^2 - 10t - 5 = 0
Dividing the equation by 5:
t^2 - 2t - 1 = 0
Using the quadratic formula:
t = (-(-2) ± √((-2)^2 - 4(1)(-1))) / (2(1))
t = (2 ± √(4 + 4)) / 2
t = (2 ± √8) / 2
t = (2 ± 2√2) / 2
t = 1 ± √2
Since time cannot be negative, we take the positive root:
t ≈ 1 + √2
Therefore, the projectile hits the ground at approximately t = 1 + √2 seconds.
c) The average rate of change in height from the time the object was launched until it hits the ground can be calculated by finding the difference in height divided by the difference in time. Let's use the known values to calculate it:
Average rate of change = (change in height) / (change in time)
Change in height = h(t) - h(0)
= -5t^2 + 10t + 5 - 5(0)^2 + 10(0) + 5
= -5t^2 + 10t
Change in time = t - 0
= t
Therefore,
Average rate of change = (-5t^2 + 10t) / t
= -5t + 10
So, the average rate of change in height is -5t + 10.
d) The velocity of the projectile can be found by taking the derivative of the height equation h(t) = -5t^2 + 10t + 5 with respect to time:
h(t) = -5t^2 + 10t + 5
h'(t) = -10t + 10
To find the velocity when the projectile is at its maximum height, we set h'(t) = 0 and solve for t:
-10t + 10 = 0
-10t = -10
t = 1
Substituting t = 1 into the derivative:
h'(1) = -10(1) + 10
h'(1) = -10 + 10
h'(1) = 0
Therefore, the velocity of the projectile when it's at its maximum height is 0. This happens because at the maximum height, the projectile momentarily stops before descending.
a) To find the height of the projectile at the moment of launch, we plug in t = 0 into the equation h(t) = -5t^2 + 10t + 5:
h(0) = -5(0^2) + 10(0) + 5
h(0) = 0 + 0 + 5
h(0) = 5
Therefore, the projectile was at a height of 5 meters at the moment of launch.
b) To find the time when the projectile hits the ground, we set h(t) equal to 0 and solve for t:
-5t^2 + 10t + 5 = 0
This is a quadratic equation, and we can solve it by factoring:
-5t^2 + 10t + 5 = (-5t - 5)(t - 1) = 0
Setting each factor equal to zero, we have:
-5t - 5 = 0 --> t = -1 (rejected because time cannot be negative)
t - 1 = 0 --> t = 1
The projectile hits the ground at t = 1 second.
c) The average rate of change in height is given by the formula:
average rate of change = (change in height) / (time)
To find the change in height from the time the object was launched until it hits the ground, we subtract the height at t = 0 from the height at t = 1:
change in height = h(1) - h(0)
Substituting the values into the equation h(t) = -5t^2 + 10t + 5:
change in height = h(1) - h(0)
= (-5(1^2) + 10(1) + 5) - (-5(0^2) + 10(0) + 5)
= (-5 + 10 + 5) - (0 + 0 + 5)
= 10 - 5
= 5
Therefore, the change in height is 5 meters.
The time it takes for the object to hit the ground is 1 second (as found in part b).
average rate of change = (change in height) / (time) = 5 / 1 = 5
Therefore, the average rate of change in height from the time the object was launched until it hits the ground is 5 meters per second.
d) The velocity of the projectile can be found by finding the derivative of the height function with respect to time.
h(t) = -5t^2 + 10t + 5
Taking the derivative:
h'(t) = -10t + 10
To find the velocity at the maximum height, we set the derivative equal to zero and solve for t:
-10t + 10 = 0
-10t = -10
t = 1
Substituting t = 1 into the derivative:
h'(1) = -10(1) + 10
h'(1) = -10 + 10
h'(1) = 0
Therefore, the velocity of the projectile when it's at its maximum height is 0.
a) To find the height of the projectile at the moment of launch, we need to substitute t = 0 into the equation.
h(0) = -5(0)^2 + 10(0) + 5
= 0 + 0 + 5
= 5
Therefore, the projectile was 5 meters high at the moment of launch.
b) To find the time at which the projectile hits the ground, we need to set the height function equal to zero and solve for t.
-5t^2 + 10t + 5 = 0
To solve this quadratic equation, we can either factor it or use the quadratic formula. In this case, it cannot be factored easily, so we will use the quadratic formula:
t = (-b ± sqrt(b^2 - 4ac)) / (2a)
For our equation, a = -5, b = 10, and c = 5. Substituting these values into the quadratic formula:
t = (-10 ± sqrt(10^2 - 4(-5)(5))) / (2(-5))
t = (-10 ± sqrt(100 + 100)) / (-10)
t = (-10 ± sqrt(200)) / (-10)
To simplify further, we can divide the numerator and denominator by -10:
t = (10 ± sqrt(200)) / 10
Now we can calculate the approximate values for t using a calculator:
t ≈ (-10 + sqrt(200)) / 10 ≈ 1.39
t ≈ (-10 - sqrt(200)) / 10 ≈ -0.89
Since time cannot be negative in this context, we discard the negative value. Therefore, the projectile hits the ground approximately 1.39 seconds after it was launched.
c) The average rate of change in height is equal to the change in height divided by the change in time. In this case, the change in height is the height at the moment of launch (5 meters) minus the height at the time the projectile hits the ground (0 meters).
Δh = 5 - 0 = 5 meters
The change in time is the time it takes for the projectile to hit the ground, which we found to be approximately 1.39 seconds.
Δt ≈ 1.39 seconds
Therefore, the average rate of change in height is:
Average rate of change = Δh / Δt
= 5 / 1.39
≈ 3.60 meters per second
So, the average rate of change in height from the time the object was launched until it hit the ground is approximately 3.60 meters per second.
d) To find the velocity of the projectile when it reaches its maximum height, we need to find the time at which the height function is maximized and then substitute this value into the derivative of the height function.
To find the time when the height function is maximized, we can use the fact that the vertex of a parabola is given by t = -b/2a. In our case, a = -5 and b = 10. Substituting these values:
t = -10 / (2(-5))
t = -10 / (-10)
t = 1
This means the projectile reaches its maximum height at t = 1 second.
Now, let's find the derivative of the height function with respect to time:
h'(t) = -10t + 10
The velocity at the maximum height is the value of the derivative at t = 1:
v = h'(1) = -10(1) + 10 = 0
Therefore, the velocity of the projectile when it's at its maximum height is 0 m/s. This makes sense because at the maximum height, the projectile momentarily stops before changing direction and falling back down.