Question

A projectile is fired straight upward from ground level with an initial velocity of 224 feet per second. At what instant will it be back at ground level? When will the height exceed 768 ft?

Answer 1

Step-by-step explanation:

We can use the kinematic equations to solve this problem. The equation we need is:

h(t) = -16t^2 + vt + h0

where h(t) is the height of the projectile at time t, v is the initial velocity, and h0 is the initial height (which is zero in this case). We can use this equation to answer both questions.

To find when the projectile is back at ground level, we need to find the time t when h(t) = 0. Substituting the given values, we get:

0 = -16t^2 + 224t

Simplifying and factoring out t, we get:

0 = t(-16t + 224)

So either t = 0 or -16t + 224 = 0. The first solution corresponds to the initial position of the projectile, so we can ignore it. The second solution gives us:

t = 14

So the projectile will be back at ground level after 14 seconds.

To find when the height exceeds 768 ft, we need to find the time t when h(t) = 768. Substituting the given values, we get:

768 = -16t^2 + 224t

Simplifying and rearranging, we get:

16t^2 - 224t + 768 = 0

Dividing by 16, we get:

t^2 - 14t + 48 = 0

Solving for t using the quadratic formula, we get:

t = (14 ± sqrt(14^2 - 4148)) / 2

t = 6 or 8

So the height of the projectile will exceed 768 ft after 6 seconds and again after 8 seconds.

Answer 2

To determine when the projectile will be back at ground level, we can use the fact that the height of the projectile can be modeled by the equation:

h(t) = -16t^2 + v0t + h0,

where h(t) is the height at time t, v0 is the initial velocity, h0 is the initial height (which is 0 in this case since it's fired from ground level), and -16t^2 is the effect of gravity.

Given:
v0 = 224 ft/s
h0 = 0 ft

To find when the projectile will be back at ground level, we need to find the time t when h(t) = 0.

0 = -16t^2 + 224t

Simplifying the equation:
16t^2 - 224t = 0

Factoring out 16t:
16t(t - 14) = 0

From this equation, we can see that either t = 0 (which is the initial time) or t - 14 = 0. However, t cannot be zero since it represents the time after the projectile is fired. Therefore, we solve for t - 14 = 0:

t - 14 = 0
t = 14

Therefore, the projectile will be back at ground level after 14 seconds.

To determine when the height exceeds 768 ft, we can set h(t) > 768 and solve for t.

-16t^2 + 224t > 768

Dividing the inequality by 16 (since it's a positive value):
-t^2 + 14t - 48 > 0

To solve this inequality, we can factor it:

-(t - 6)(t - 8) > 0

Now, we need to find the intervals where this inequality is true. The product of two factors is positive when both factors have the same sign (both positive or both negative).

Case 1: (t - 6) > 0 and (t - 8) > 0
t > 6 and t > 8

In this case, t must be greater than 8 since it satisfies both conditions.

Case 2: (t - 6) < 0 and (t - 8) < 0
t < 6 and t < 8

In this case, t must be less than 6 since it satisfies both conditions.

Therefore, the solution to the inequality is t < 6 or t > 8.

This means the height of the projectile will exceed 768 ft at some time before 6 seconds or after 8 seconds.

To summarize:

The projectile will be back at ground level after 14 seconds.
The height of the projectile will exceed 768 ft before 6 seconds or after 8 seconds.