Question

1. projectile recall that the acceleration due to earth’s gravity is 32 ft/sec². from ground level, a projectile is fired straight upward with velocity 90 feet per second. (a) what is its velocity after 3 seconds? (b) when does it hit the ground? (c) when it hits the ground, what is the net distance it has traveled? (d) when it hits the ground, what is the total distance it has traveled?

Explanation:

Part (a)

Step1: Recall velocity formula

For vertical motion under gravity, the velocity function is \( v(t) = v_0 - gt \), where \( v_0 \) is initial velocity, \( g \) is acceleration due to gravity. Here, \( v_0 = 90 \) ft/s, \( g = 32 \) ft/s², \( t = 3 \) s.

Step2: Substitute values

\( v(3)=90 - 32\times3=90 - 96=- 6 \) ft/s. The negative sign indicates direction is downward.

Step1: Recall position function

The position function for vertical motion is \( s(t)=v_0t-\frac{1}{2}gt^2 \). When it hits the ground, \( s(t) = 0 \) (since it starts from ground level, \( s(0)=0 \)). So \( 0 = 90t-\frac{1}{2}\times32t^2 \).

Step2: Solve the equation

Simplify: \( 0 = 90t - 16t^2 \), factor out \( t \): \( t(90 - 16t)=0 \). Solutions are \( t = 0 \) (initial time) and \( 90 - 16t=0\Rightarrow t=\frac{90}{16}=\frac{45}{8} = 5.625 \) s.

Step1: Understand net distance

Net distance is the displacement, which is \( s(t_2)-s(t_1) \). Since it starts at \( s(0) = 0 \) and ends at \( s(t)=0 \) (when it hits ground), net distance \( = 0 - 0=0 \) ft.

Answer:

-6 ft/s

Part (b)