The height in meters of a projectile launched from the top of a building is given by h(t)=-5t^2+60t+15, where t is time in seconds since it was launched.
a) How tall is the building?
b) At what time does the projectile hit the ground?
C)What is the velocity(rate of change with respect to time) of the object when it hits the ground?
d)Determine the projectile's velocity at 2 seconds.
YOU have gravity reversed !!!!
the building is 15 meters high since that is h when t = 0
h = -5 t^2 + 60 t + 15
at ground
0 = -5 t^2 + 60 t + 15
t^2 - 12 t - 3 = 0
t^2 -12 t = 3
t^2 - 12 t + 36 = 39
(t-6)^2 = 39
t - 6 = +/- sqrt 39
t = 6 + 6.24 = 12.24 seconds
dh/dt = -10 t + 60
= -10(12.24) + 60
= -62.4 m/s
dh/dt =-10 t + 60 at 2 seconds
= -20 + 60
= 40 m/s
Thank you Damon :)
To answer these questions, we need to analyze the given equation and its properties. Recall that the height of the projectile at any time t is given by the equation h(t) = -5t^2 + 60t + 15.
a) To find the height of the building, we need to determine the maximum value of the quadratic equation, as this represents the height of the projectile at its highest point. We can identify the maximum point of the quadratic equation using calculus or by considering its symmetry. In this case, we'll use calculus.
To find the maximum point, we need to take the derivative of the equation, which represents the rate of change of the height with respect to time: h'(t) = -10t + 60.
Setting h'(t) equal to zero, we get: -10t + 60 = 0.
Solving for t, we find that t = 6.
Substituting t = 6 back into the original equation, we get h(6) = -5(6)^2 + 60(6) + 15 = 195.
Therefore, the height of the building is 195 meters.
b) To determine the time at which the projectile hits the ground, we need to find when the height of the projectile becomes zero. We can set the height equation h(t) = 0 and solve for t.
-5t^2 + 60t + 15 = 0.
To solve this quadratic equation, we can use the quadratic formula: t = (-b ± √(b^2 - 4ac)) / 2a.
Plugging in the values a = -5, b = 60, and c = 15 into the quadratic formula, we get:
t = (-60 ± √(60^2 - 4(-5)(15))) / (2(-5)).
Simplifying further gives:
t = (-60 ± √(3600 + 300)) / (-10).
t = (-60 ± √3900) / (-10).
We can continue to simplify this expression:
t = (-60 ± 62.48) / (-10).
This gives two possible values for t:
t1 = (-60 + 62.48) / (-10) ≈ 0.245 seconds.
t2 = (-60 - 62.48) / (-10) ≈ 12.45 seconds.
Since the question asks for the time it hits the ground, we discard t1 as it represents an earlier time when the projectile hasn't yet been launched. Therefore, the projectile hits the ground at approximately 12.45 seconds.
c) To find the velocity (rate of change of height with respect to time) when the projectile hits the ground, we can take the derivative of the height equation h'(t) = -10t + 60 and substitute the time t when the projectile hits the ground, which is approximately 12.45 seconds:
h'(12.45) = -10(12.45) + 60 ≈ -124.5 + 60 ≈ -64.5 m/s.
Therefore, the velocity of the object when it hits the ground is approximately -64.5 m/s, where the negative sign indicates that the motion is downward.
d) To determine the projectile's velocity at 2 seconds, we can substitute t = 2 into the derivative of the height equation:
h'(2) = -10(2) + 60 = -20 + 60 = 40 m/s.
Therefore, the projectile's velocity at 2 seconds is 40 m/s.