A projectile is shot upwards with an initial velocity of 30 m/s. Its height at time t is given by h=30t-4.9t^2. During what period is the projectile more than 40 m above the ground?

Put your height h as 40. Well, for any time t, the height is given by 30t - 4.9t^2 = h.

So, since you height has to be above 40, the inequality becomes 30t - 4.9t^2 > 40, higher than 40 m

Now that you have 30t - 4.9t^2>40, can you solve for t? If, not, post and I or someone else will guide you.

Height of projectile above ground = h

h(t) = h as a function of t, t = time.

h(t) = 30t - 4.9t^2

Question: When is the projectile more than 40 meters above ground?

Answer:

Projectile is 40 meters above the ground

h(t) > 40

30t - 4.9t^2 > 40

Factor out t:

t(30 - 4.9t) > 40

Height of projectile above ground = h

h(t) = h as a function of t, t = time.

h(t) = 30t - 4.9t^2

Question: When is the projectile more than 40 meters above ground?

Answer:

Projectile is 40 meters above the ground implies that the following are true...

h(t) > 40

30t - 4.9t^2 > 40

-4.9t^2 + 30t - 40 > 0

Find roots of -4.9t^2 + 30t - 40:

a = -4.9
b = 30
c = -40

t = [-b (+/-) sqrt(b^2 - 4ac)] / 2a = [-30 (+/-) sqrt(900 - 4(-4.9)(-40))] / 2(-4.9)

We get the roots:

t = 1.9622 (first root)
t = 4.1602 (second root)

All we have done is find the values of t that make h(t) - 40 = 0.

The projectile is more than 40 meters above the ground between 1.9266 seconds and 4.1602 seconds after it is shot.