Initial velocity Suppose a baseball is thrown vertically upwar...

Question

Initial velocity Suppose a baseball is thrown vertically upward from the ground with an initial velocity of v0ft/s Its height above the ground after t seconds is given by s(t) = -16t²+v0t. Determine the initial velocity of the ball if it reaches a high point of 128 ft.

Accepted Answer

Welcome back, everyone. A basketball is shot vertically upward with an initial velocity of V knot feet per second from a player's hand. The height of the basketball above the ground after T seconds is given by S of T equals -16 T2 plus V 0 T. If the basketball reaches a peak height of 64 ft, what was the initial velocity? A 128 ft per second, B 16 ft per second, C 32 ft per second, and D 64 ft per second. So in this context of the problem, we're looking at the peak height and we're given the height function, it says S of T. So we can essentially find the derivative of S of T, which means that we are differentiating -16 t2 plus V not T, right? The reason why we're looking for the derivative is because at the critical point we might have a maximum or a minimum. So let's identify the derivative. The derivative of T2d based on the power rule if it is to T, so we get negative 32 T. B 0 is a constant and the derivative of T is 1, so we get plus V0 and this is also our velocity function. So now what we want to do is simply set it equal to. C, right? The reason why we set it to 0 is because at the critical point where SFT is equal to 0, we might have a minimum or a maximum. So -302 T plus V is equal to 0, and we can say that V0 is equal to 302 T. Now, first of all, we want to prove that this is a point of a maximum, and if we look at the given function, it is a parabola that opens down because the leading coefficient is negative, right? So essentially we have a problem that opens down. We already understand that at The critical point we will have a horizontal tangent line, and that critical point actually corresponds to the vertex of the parabola because this is where we have a horizontal tangent line. So indeed we can show that the not is a point of maximum because the leading coefficient. Is negative. It's -16, right? So we have shown that we have a point of maximum height. And now what we can do is simply substitute it into the equation S of T. So Of tea Is equal to 64. And this corresponds to the maximum, right? So we have 64 ft. What we're going to do is simply set 64 equal to -16T. Squared Plus be not T. And now we know that V is equal to 32T. So we can substitute that and we're going to get 64 equals -16 T squared. Plus 32 T multiplied by T gives us 32 T squared. What we want to do is simply solve for the time value. We can see that 64 is equal to 16 T2d when we combine the terms. And now dividing both sides by 16, we can see that T2d is equal to 4. And since time is not negative, we can take square root of both sides, so time is equal to 2. Now that we have our time value, we can identify the not. We know that the knot is 32 T. Right, so our time value is 2. So we get 32 multiplied by 2, and this means that the 0 is equal to 64. Pete per second. We have used our previous equation. We have located the time at which we reached the maximum height, and we went back and substituted the time for the maximum height to identify the initial velocity of 64 ft per second, meaning the final answer to this problem corresponds to the answer choice D. Thank you for watching.

Initial velocity Suppose a baseball is thrown vertically upward from the ground with an initial velocity of v0ft/s Its height above the ground after t seconds is given by s(t) = -16t²+v0t. Determine the initial velocity of the ball if it reaches a high point of 128 ft.

Key Concepts

Quadratic Functions

Vertex of a Parabola

Initial Velocity

Watch next