An object thrown in the air on a planet in a distant galaxy is at height
s(t) = -20t^2 +67t +42 feet at time t seconds after it is thrown.
a) From what height was the object thrown?
b) With what velocity was the object thrown?
c) What is the acceleration due to gravity on this planet?
To find the answers to these questions, we will need to analyze the given equation for the height of the object as a function of time.
a) From what height was the object thrown?
To find the height at which the object was throw, we can look at the equation s(t) = -20t^2 + 67t + 42. The constant term represents the initial height at t=0, which is when the object was thrown. Therefore, the initial height of the object can be found by evaluating the equation at t = 0:
s(0) = -20(0)^2 + 67(0) + 42
s(0) = 42
So, the object was thrown from a height of 42 feet.
b) With what velocity was the object thrown?
The velocity of the object can be found by taking the derivative of the height function, s(t), with respect to time, t. Differentiating the function will give us the rate of change of height with respect to time, which is the velocity.
s'(t) = -40t + 67
At the time of throwing the object, t = 0, we can find the initial velocity by evaluating the velocity function:
s'(0) = -40(0) + 67
s'(0) = 67
Therefore, the object was thrown with an initial velocity of 67 feet per second.
c) What is the acceleration due to gravity on this planet?
The acceleration due to gravity can be determined from the quadratic equation, s(t) = -20t^2 + 67t + 42. In this equation, the coefficient of the t^2 term represents half of the acceleration due to gravity.
Therefore, the acceleration due to gravity on this planet would be:
a = -20/2
a = -10
Hence, the acceleration due to gravity on this planet is -10 feet per second squared.
a) sub in t=0 into s(t)
b) find the derivative to get velocity
c) find the derivative of velocity to get acceleration.
You will get a constant.