How long will it take to hit the ground?

Question

Suppose you throw a baseball straight up at a velocity of 64 feet per second. A function can be created by expressing distance above the ground, s, as a function of time, t. This function is s = -16t2 + v0t + s0
•16 represents 1/2g, the gravitational pull due to gravity (measured in feet per second2).
•v0 is the initial velocity (how hard do you throw the object, measured in feet per second).
•s0 is the initial distance above ground (in feet). If you are standing on the ground, then s0 = 0

Answer 1

I think you can use a simpler equation than the one for distance.
We don't need to find distance, just time.

v = u + a.t
v= 0, u= 64 f/s, a= -32f/s^2
solve for t.
t = 2 s
Thats the time it takes to go up just before it starts to fall again.
The time taken to fall is going to be the same: another 2 s

total time: 4 s

Answer 2

Velocity = Initial Velocity + (Acceleration * Time)

We know that the velocity at the vertex (highest point) = 0

0 f/s = 64 f/s + -32 f/s^2 (t)

so to get to the vertex takes -64 f/s divided by -32f/s^2 = 2 s

the acceleration is the same throughout so to get back to the ground will take another 2 secs.

for a total = 4 seconds

(that is based on the assumption that the terminal velocity of the object is greater than 64 ft/s)

Answer 3

Given
u = Initial velocity with which the ball is thrown up = 64'/sec
v = Final velocity the ball attained after reaching the maximum height (s) = 0'/sec
t = time taken to reach the maximum height.
g = acceleration due to gravity = -32'/sec^2
s=Vertical distance travelled by base ball = ?
Applying the equation v = u+at, we get
0 = 64 - 32t
Or t = 2sec
Then applying the equation s = ut +1/2 gt^2
s=64 x 2 - 0.5 x 32 x 2^2 = 64'
That means the baseball takes 4 secs to reach back the starting point after raching a maximum height of 64'. And total distance travelled is 128'.

Answer 4

another way to solve this problem using calculus and differentiation...
v(t) is the derivative of s(t)

s(t)=-16t^2 + v0t + s0
s(t)=-16t^2 + 64t
v(t)=-32t + 64
0=-32t + 64
32t=64
t=2

as the previous person said, the time going up is the same as the time going down, so double t and you get a time of 4 seconds.

Answer 5

7 seconds

Answer 6

s = -16t^2 + 64t

s = 0 when t = 0, and time when baseball hits the ground

so, -16t^2 + 64t = 0
-16t (t - 4t) = 0
t = 0 or t = 4

so, it takes 4 seconds to hit the ground

Answer 7

set the equation I gave you s=-16t^2+64t=0 and solve for t. I'll do "most" of the work for you:

-16t^2+64t=16t(4-t)=0.

Answer 8

4 secs lol

i think the equation v=u+at is the bttr choice 4 this Q
it was a mins play with it

Answer 9

I think this cannot be solved dear until u know the initial acceleration

Answer 10

s (distance) = 1/2 g(gravity) x t2 (time squared)