Okay I have :
a ball is thrown with an initial velocity of 40 ft/second from a height of 4 feet. h(t)=-16t+40t+4
a) what is the highest point reached by the ball?
b) at what time will the ball reach the highest point?
your formula is wrong.
h(t)=-16t^2 + 40t+4
v(t)=40-16t
so max height is when v(t)=0 or t=40/16
and the max height is found by plugging that into the height equation.
5 seconds
To find the highest point reached by the ball, we need to determine the vertex of the trajectory function h(t) = -16t^2 + 40t + 4.
a) To find the highest point, we can use the formula for the x-coordinate of the vertex, which is given by t = -b/2a, where a is the coefficient of the t^2 term and b is the coefficient of the t term. In this case, a = -16 and b = 40. Substituting these values into the formula, we have:
t = -40 / (2 * -16)
t = 40 / 32
t = 5/4
t = 1.25 seconds
Now we can find the height of the ball at this time by substituting t = 1.25 into the h(t) equation:
h(1.25) = -16(1.25)^2 + 40(1.25) + 4
h(1.25) = -16(1.5625) + 50 + 4
h(1.25) = -25 + 50 + 4
h(1.25) = 29 feet
Therefore, the highest point reached by the ball is at a height of 29 feet.
b) The time at which the ball reaches the highest point is 1.25 seconds.