Myrtle throws a ball upward from a second-floor balcony, 6 m above the ground with an initial velocity of 2 m/s. In this situation, u = 2 and ho = 6, so the relation that models the height of the ball is h = -5t2 + 2t + 6. Myrtle knows that changing the velocity with which she throws the ball will change the maximum height of the ball. Myrtle wants to know with what velocity she must throw the ball to make it pass over a tree that is 11 m tall.
a) Suppose that Myrtle just dropped the ball from the balcony, with an initial velocity of 0 m/s Write the quadratic relation that models this situation.
b) What is the maximum height of the ball?
V^2 = Vo^2 + 2g*h
Vo^2 = V^2-2g*h = 0 + 20*(11-6) = 100
Vo = 10 m/s. To pass over tree.
a. 0.5g*t^2 = 6
b. h max = 6 m. = Ht. of balcony.
a) When Myrtle drops the ball from the balcony, the initial velocity is 0 m/s. In this case, the equation that models the height of the ball is obtained by setting the initial velocity (u) in the original equation to 0:
h = -5t^2 + 2t + 6 (where u=0 and ho=6)
b) To find the maximum height of the ball, we need to determine the vertex of the quadratic equation. The vertex of a quadratic equation is given by the formula t = -b/2a, where a and b are the coefficients of the quadratic equation.
In this case, the equation that models the height of the ball is h = -5t^2 + 2t + 6. Comparing this to the standard form of a quadratic equation (ax^2 + bx + c = 0), we can see that a = -5 and b = 2.
Using the formula t = -b/2a, we can calculate the t-coordinate of the vertex as follows:
t = -2/(2*(-5))
t = -2/-10
t = 0.2 seconds
To find the maximum height, substitute the value of t back into the equation:
h = -5(0.2)^2 + 2(0.2) + 6
h = -5(0.04) + 0.4 + 6
h = -0.2 + 0.4 + 6
h = 6.2 meters
Therefore, the maximum height of the ball when dropped from the balcony is 6.2 meters.