Question

A projectile is launched from ground level with an initial velocity of V₀ ft./s. Neglecting air resistance, its height in feet, t seconds after launch is given by S = -16t^2 + V₀. Find the time(s) that the projectile will:

(1) Reach a height of 160 feet
(2) Return to the ground when V₀ = 112 ft./s
a) Solve for t when S = 160
b) Solve for t when S = 0
c) Solve for t when V₀ = 160
d) Solve for t when S = -112

Answer 1

To find the times at which the projectile reaches a particular height or returns to the ground, we solve a quadratic equation representing the vertical motion of the projectile. By substituting the specific height values in the equation, we can solve for the time(s) when these events occur.

Step-by-step explanation:

The equation of motion for a projectile launched from the ground can be written as S = -16t^2 + V0t, where S is the height in feet and t is the time in seconds. When solving for the time t that the projectile reaches a certain height or returns to the ground, we set the corresponding value of S and solve the quadratic equation.

Discussion for (a)

When S = 160 feet, the quadratic equation is -16t^2 + 112t - 160 = 0. Solving this gives the time(s) at which the projectile reaches 160 feet.

Solution for (b)

When the projectile returns to the ground, S = 0. The equation becomes -16t^2 + 112t = 0, and solving it provides the time(s) the projectile takes to hit the ground.

Parts (c) and (d) involve finding the times when V0 = 160 ft./s and when S = -112 feet, which would not normally occur for a projectile launched from ground level and is therefore likely a misunderstanding of the problem parameters.