Math help?

Question

a projectile is launched upward from ground level with an initial speed of 98 m/s. how high will it go? when will it return to the ground?

Answer 1

The height, in meters, at time t, with t in seconds, is

h(t) = -(1/2)gt^2 + v0*t + s0

where s0 = initial height, in this case 0
v0 = initial velocity, in this case 98
g = gravity constant, 9.8 m/s^2

therefore h(t) = -4.9t^2 + 98t.

to find the maximum height, complete the square and rewrite h(t) as

h(t) = -4.9(t^2 - 20t) = -4.9[(t-10)^2 -100] =
-4.9(t-10)^2 + 490

using the fact that (t-a)^2 = t^2 - 2at + a^2. since we want -2at = -20t, it means a=10.

-4.9(t-10)^2 <= 0 for all t, and it's = 0 when t=10. Therefore the maximum height is 490 m.

To determine the time at which it hits the ground, set h(t) = 0 and solve for t.

You have -4.9t^2 + 98t = 0, or -4.9t(t-20)=0. Therefore t=0 or t=20. t=0 is when it's launched, so it hits the ground when t=20.

Answer 2

Gravity is -9.8 m/s^2. So, It has an initial velocity of 98 m/s, a final velocity (at its peak) of 0 m/s, and an acceleration of -9.8 m/s^2. Simply plug into a formula from there. (For time, v final = v initial + at, so 0 = 98 + -9.8(t), meaning t = 10 to reach the peak, 10 more to fall down, so 20 seconds).

Answer 3

acceleration due to gravity is -9.8m/s/s
it will travel upwards for 10 seconds, average velocity of 49m/s
maximum height 490 meters.
touchdown total of 20 seconds later.

Answer 4

490 m
20 sec

Answer 5

http://galileo.phys.virginia.edu/classes/109N/more_stuff/Applets/ProjectileMotion/jarapplet.html

Just enter your speed and angle of 90 degrees to get a maximum height of 489.47 m and a time of 19.98 s.