Question

A projectile is fired straight up from an initial height of 28 ft. above the ground with an initial velocity of 76 ft/sec. For what time period is the object at least 88 ft above the ground?

a) 1/2 seconds
b) 2 seconds
c) 3/2 seconds
d) 3 seconds

Answer 1

To determine the time period for which the object is at least 88 ft above the ground, we can use the equations of motion for vertical motion. The object is at least 88 ft above the ground for a time period of 1/2 seconds to 3 seconds.

Step-by-step explanation:

To determine the time period for which the object is at least 88 ft above the ground, we can use the equations of motion for vertical motion.

The equation for the height of a projectile at any time is given by:

h(t) = h₀ + v₀t - 16t²

Where:

• h(t) is the height of the projectile at time t
• h₀ is the initial height of the projectile
• v₀ is the initial velocity of the projectile
• t is the time elapsed

Substituting the given values:

h(t) = 28 + 76t - 16t²

To find the time period for which the object is at least 88 ft above the ground, we need to solve the equation:

88 = 28 + 76t - 16t²

Simplifying the equation, we get:

-16t² + 76t + 60 = 0

Using the quadratic formula, we find that t ≈ 1/2 seconds or t ≈ 3 seconds.

Therefore, the object is at least 88 ft above the ground for a time period of 1/2 seconds to 3 seconds.