TchrWill ... the equation is correct as written.
Since the units for height are in meters, the acceleration due to gravity is being approximated as 10 m/s<sup>2</sup> ... therefore the coefficient of -5 for the quadratic term.
kayla_t99 said:
A ball is thrown into the air from a building and falls to the ground below. The height of the ball, h metres, relative to the ground t seconds after being thrown is given by h=-5t²+30t+35.
a) determine the maximum height of the ball above the ground.
b) how long does it take the ball to reach the maximum height?
c) after how many seconds doe the ball hit the ground?
h = -5t<sup>2</sup> + 30t + 35 is a quadratic equation whose graph is an inverted parabola.
the vertex of the parabola will give you the time when it reaches its max height, and using that value for t, one can find the max height.
for a quadratic of the form y = ax<sup>2</sup> + bx + c, the vertex is located at
x = -b/(2a) ...
for your problem, the vertex is located at t = -30/[2(-5)] = 3 seconds ... this is the time that the projectile reaches its maximum height.
so ... substitute 3 seconds for t in the height equation to get the max height of the projectile
h<sub>max</sub> = -5(3)<sup>2</sup> + 30(3) + 35 = 80
meters
to find the time the projectile hits the ground, set h = 0 and solve for t ...
-5t<sup>2</sup> + 30t + 35 = 0
divide each term by -5 ...
t<sup>2</sup> - 6t - 7 = 0
factor ...
(t - 7)(t + 1) = 0
so the projectile hits the ground at either t = 7 seconds or t = -1 seconds ... which solution makes sense in the context of the problem?