the graph shows the height of a projectile t seconds after it is launched. if acceleration due to gravity is…

Explanation:

Step1: Recall the projectile motion formula

The formula for the height of a projectile is ( h(t) = at^2 + vt + h_0 ), where ( a ) is the acceleration due to gravity, ( v ) is the initial velocity, and ( h_0 ) is the initial height. We know ( a=-16\space ft/s^2 ).

Step2: Analyze the initial conditions

When ( t = 0 ) (at launch), the height ( h(0) ) is the initial height ( h_0 ). Also, if the projectile is launched vertically (or with some horizontal component, but the formula still holds for vertical height), at the peak of the motion, the velocity ( v(t) = v+2at = 0 ). But another way: if we assume the graph shows that at ( t = 0 ), the height is 5? Wait, no, wait. Wait, maybe the graph (even though not shown, but from the options) - wait, the first option is ( h(t)=-16t^2 + 5 ), second ( -16t^2+32 ), third ( -16t^2 + 42t + 5 ). Wait, but let's think about initial velocity. If the projectile is launched, and if at ( t = 0 ), the height is ( h_0 ), and if there's no initial vertical velocity (maybe dropped? No, launched). Wait, but the formula ( h(t)=at^2 + vt + h_0 ), with ( a=-16 ). If the projectile is launched from a height of 5 (so ( h_0 = 5 )) and maybe initial velocity? Wait, no, maybe the graph shows that when ( t = 0 ), ( h(0)=5 ), so ( h_0 = 5 ). Then, let's check the options. The third option is ( h(t)=-16t^2 + 42t + 5 ), which has ( h_0 = 5 ), and initial velocity ( v = 42 ). But wait, maybe the graph shows that at ( t = 0 ), height is 5, and maybe the motion. Wait, but maybe the first option: ( h(t)=-16t^2 + 5 ) would mean initial velocity ( v = 0 ) (dropped from height 5). The second option: ( h(t)=-16t^2 + 32 ) has ( h_0 = 32 ), ( v = 0 ). The third has ( h_0 = 5 ), ( v = 42 ). But since the acceleration is -16, and the formula is ( h(t)=at^2 + vt + h_0 ), so ( a=-16 ). Now, if the projectile is launched (not dropped), there should be an initial velocity. But maybe the graph (not shown) has ( h(0)=5 ), so ( h_0 = 5 ), so the third option is ( h(t)=-16t^2 + 42t + 5 ), which has ( h_0 = 5 ), and initial velocity ( v = 42 ). But wait, maybe the graph shows that at ( t = 0 ), height is 5, so ( h_0 = 5 ), so the third option is correct? Wait, no, wait. Wait, maybe I made a mistake. Wait, the formula for projectile motion: the vertical height is ( h(t) = -16t^2 + vt + h_0 ), where ( v ) is initial vertical velocity, ( h_0 ) initial height. So if the projectile is launched, there must be an initial vertical velocity (unless dropped, ( v = 0 )). So if the graph shows that at ( t = 0 ), height is 5, then ( h_0 = 5 ), so the third option ( h(t)=-16t^2 + 42t + 5 ) has ( h_0 = 5 ), so that's the one with initial height 5 and initial velocity 42. So the correct equation should be ( h(t)=-16t^2 + 42t + 5 ).

Answer:

( h(t) = -16t^2 + 42t + 5 ) (the third option: ( h(t) = -16t^2 + 42t + 5 ))