Vanessa throws a tennis ball in the air. The function h(t)=-16t^2+45t+7 represents the distance, in feet, that the ball is form the grounf at any time t. At what time, to the nearest tenth of a second, is the ball at its maximum height?
If the function h(t)= +45t -16t^2 +7 represents the distance, in feet, that the ball is form the ground at any time,the ball is thrown upward with a velocity of 45fps from a height of 7 feet above the ground.
The final zero velocity of zero is reached from Vf = Vo - gt or 0 = 45 - 32t making the maximum height time 1.406 seconds or 1.4 seconds.
Alternatively, taking the first derivitive of h = Vot - (gt^2)/2 or dh/dt = 45t - 32t = 0 making t = 1.4 sec.
the answer is 2.86 when rounded to the nearest hundred
To find the time when the ball is at its maximum height, we need to determine the vertex of the parabolic function h(t) = -16t^2 + 45t + 7.
The vertex of a parabola in the form of y = ax^2 + bx + c can be found using the formula:
t = -b / (2a)
For our function h(t) = -16t^2 + 45t + 7, the coefficient of t^2 is -16 and the coefficient of t is 45. Plugging these values into the formula, we get:
t = -(45) / (2 * (-16))
t = -45 / -32
t ≈ 1.40625
Rounding to the nearest tenth of a second, the ball is at its maximum height at approximately 1.4 seconds.
To find the time at which the ball is at its maximum height, we need to determine the vertex of the quadratic function h(t) = -16t^2 + 45t + 7. The vertex represents the highest point of the graph, which corresponds to the maximum height of the ball.
The vertex of a quadratic function in the form h(t) = at^2 + bt + c can be found using the formula t = -b / (2a). In our case, the equation is h(t) = -16t^2 + 45t + 7, so a = -16 and b = 45.
Substituting these values into the formula, we get t = -45 / (2 * -16). Simplifying further, we find t = -45 / -32. This can be rewritten as t = 1.40625.
Since we are asked to round to the nearest tenth of a second, the time at which the ball reaches its maximum height is approximately 1.4 seconds.