Projectile's Height

[nellkoda]

A projectile's height (h), in ft., as a function of time (t), in seconds after launch, is h(t)=-16t^2+80t

a. At what times will the projectile be on the ground?

b. At what time will the projectile be at its maximum height?

c. What is the projectiles maximum height?

d. How long after launch will the projectile be 48ft in the air?

 

[skeeter]

A projectile's height (h), in ft., as a function of time (t), in seconds after launch, is h(t)=-16t^2+80t

a. At what times will the projectile be on the ground?

set h(t) = 0 and solve for t ... you'll get two solutions

b. At what time will the projectile be at its maximum height?

max height will be the vertex of the parabola formed by the graph of h(t) ... does -b/(2a) ring a bell?

c. What is the projectiles maximum height?

max height is h[-b/(2a)]

d. How long after launch will the projectile be 48ft in the air?

set h(t) = 48 and solve for t ... once again you'll get two solutions for the time

 

[Mrspi]

A projectile's height (h), in ft., as a function of time (t), in seconds after launch, is h(t)=-16t^2+80t

a. At what times will the projectile be on the ground?

b. At what time will the projectile be at its maximum height?

c. What is the projectiles maximum height?

d. How long after launch will the projectile be 48ft in the air?

a) The projectile will be on the ground when the height, h, is 0. Substitute 0 for h, and solve the resulting quadratic equation for t.

b) The graph of h = -16t^2 + 80t is a parabola which opens downward. The maximum height occurs at the vertex of the parabola, and the time at which the maximum height is reached is the value of t at the vertex.

c) The maximum height reached by the projectile is the value of h at the vertex of the parabola.

d) To find the time(s) at which the height of the projectile is 48 feet, substitute 48 for h and solve the resulting quadratic equation for t.

Now, you see if you can find the answers to the questions, following the suggestions I've given for each part. If you are still having trouble, please repost showing all of your steps (even if you think they are wrong) so we can see where in the process your difficulties lie.

 

[TchrWill]

A projectile's height (h), in ft., as a function of time (t), in seconds after launch, is h(t)=-16t^2+80t

a. At what times will the projectile be on the ground?
b. At what time will the projectile be at its maximum height?
c. What is the projectiles maximum height?
d. How long after launch will the projectile be 48ft in the air?

Taking it step by step:

1--Vf = Vo - gt where Vo = initital velocity in fps, Vf = final velocity in fps, t = time in seconds and g = acceleration due to gravity, 32fps^2.
2--The projectile is on the ground at time t = 0 obviously.
3--With Vo = 80 fps, and Vf = 0, 0 = 80 - 32t making the time to its maximum height t = 2.5 seconds.
4--The maximum height derives from the expression you gave, h = Vot - gt^2 (in general form) where h = 80(2.5) - 16(2.5^2) = 100 ft.
5--The projectile will be at a height of 48 ft. twice during the flight; once on the way up and once on the way down. On the way up, 48 = 80t - 16t^2, a quadsatic that yields h = .7 seconds on the way up and 4.3 seconds on the way down.
6--The time to 48 frrt height on the way down could also be derived from 2.5 sec. to zero velocity (maximum height) up plus (100 - 48) = 0(t) + 16t^2 down to 48 ft. or 2.5 + t(down) = 1.8 = 4.3 seconds total.