# Thread: ball is thrown into the air; max height + seconds.

1. ## ball is thrown into the air; max height + seconds.

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=-5tē+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?

2. 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=-5tē+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?
If you are on earth, your expression should be h = 30t - 16t^2 + 35 assuming the ball is thrown from atop a 35 high building, your initial upward, and vertical velocity is 30 (no units are given) and gravity is 32 ft./sec.^2.

The time to the maximum height derives from Vf = Vo - gt where Vf = the final velocity, 0, Vo = the initial velocity (not given), g = the acceleration due to gravity and t = the time.

Therefore, 0 = Vo - 32t from which you can get t.

Having t and the correct formula, you can find h(max) above the ground.

Having h(max), the time to fall back to the ground derives from h(max) = o(t) + 16t^2 or.

Add the two times together foro the total time of flight.

You have only to define the units of your initial upward velocity, which are more than likely ft./sec..

3. ## Re: ball is thrown into the air; max height + seconds.

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.

Originally Posted by kayla_t99
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=-5tē+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?

4. ## Re: ball is thrown into the air; max height + seconds.

Originally Posted by skeeter
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.
Many thanks for the heads up alert. Been so used to using ft/sec. for 55 years that it just never crossed my mind.

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