I am stuck on three algebra questions can anyone assist me ?

Question

I am just really pressed for time to get all this work done and my little girl is down with a high fever. I just need some help I cannot figure out for the life of me what they are and I am so beat.


A ball is thrown vertically upward, from the top of a building 80 m high with an initial velocity of 10 m/s. After t s, the height h is given by
When is the ball at a height of 65 m?



A ball is thrown vertically upward, from the top of a building 30 m high with an initial velocity of 5 m/s.
After t s, the height h is given by
When does the ball hit the ground?



A ball is thrown vertically upward, from the top of a building 30 m high with an initial velocity of 5 m/s. After t s, the height h is given by
When does the ball hit the ground?

Answer 1

Your equations are all:

h = -1/2 a * t^2 + v0 * t + s

... where:
a is acceleration due to gravity,
v0 is the initial velocity upward, and
s is the initial height.

(1) A ball is thrown vertically upward, from the top of a building 80 m high with an initial velocity of 10 m/s

Assuming you're using 10 m/s^2 for acceleration (a fairly close approximation to reality)

h = -1/2 a * t^2 + v0 * t + s
h = -1/2 * 10 * t^2 + 10 * t + 80
h = -5 * t^2 + 10 t + 80

If you want to solve for h=65m:

h = -5 * t^2 + 10 t + 80
65 = -5 * t^2 + 10t + 80
5t^2 - 10t - 15 = 0

Divide by 5:

t^2 - 2t - 3 = 0
(t + 1)(t - 3) = 0

t = -1 seconds, t = 3 seconds. The negative answer can be ignored, so the answer you want is 3 seconds.

===================
(2) A ball is thrown vertically upward, from the top of a building 30 m high with an initial velocity of 5 m/s.
After t s, the height h is given by
When does the ball hit the ground?

h = -1/2 a * t^2 + v0 * t + s
h = -1/2 * 10 * t^2 + 5 * t + 30
h = -5 t^2 + 5t + 30

Solve for h=0:

h = -5 t^2 + 5t + 30
5 t^2 - 5t - 30 = 0
t^2 - t - 6 = 0
(t + 2)(t -3) = 0

t = -2 (ignored), t = 3. 3 seconds until it hits the ground.

(Your third question is identical to the second.)

Answer 2

I think the overarching concept that the educators who asked these questions to you are trying to get across to you is that the average speed of an object projected in to the air for the entire distance traveled is 1/2 of the initial speed (if less than terminal velocity) because the full speed is calculated and launch and a speed of 0 is calculated at the point where to object ends its ascent and begins its decent.

Answer 3

1. h = initial_height + ut + 0.5 a t^2

where u=initial velocity,
a= acceleration (deceleration) due to gravity = -9.7536 m/sec/sec .
NOTE that a is negative here, because gravity is acting against the direction in which the ball is thrown

So h = 80 + 10t - 0.5 (9.7536) t^2
Plug in h= 65 above, you will get a quadratic equation in t. Solve it; you will probably get one t value that is < 0 (dump it), and one that's > 0

Q2. Similar logic....
h = 30 + 5t - 0.5 (9.7536) t^2
Plug in h=0, solve for t