Barry M. answered • 01/16/20

Professor, CalTech Grad; Many Years Tutoring Math, SAT/ACT Prep, Chem

Assuming the projectile is launched vertically from the ground (or even if launched at an angle, the vertical component of velocity) at a velocity of V_{0} ft/sec, and on the earth gravity is pulling down at 32 ft/sec^{2},

the velocity in feet/sec will be expressed as

v = V_{0} - 32T, and

the height in feet will be expressed as

f = V_{0}T - 16T^{2}

a. V_{0}T - 16X4 = V_{0}T - 64

b. V_{0}T - 16X49 = V_{0}T - 784

c. The maximum height will occur when the projectile has lost its upward velocity and hence v = 0. So,

V_{0}- 32T = 0, and T = V_{0}/32.

Plugging into the above equation for f,

f = V_{0}(V_{0}/32) - 16 (V_{0}/32)^{2}

= V_{0}^{2}(1/32 - 1/64)

= V_{0}^{2}/64

d. Plugging again into the top equation for f, and setting f = 0 and factoring gives

T(V_{0} - 16T) = 0. Solve for T. So f will = 0 at T=0, which was the moment of launch, and not the time when it will hit the ground. The answer will then be the other root of this equation. It follows that

(V_{0} - 16T) = 0, so T = V_{0}/16. Note that the equation for f is a quadratic in T, and graphs would show it's symmetric. The time ascending is = time descending, so the projectile hits the ground at double the time it took to reach the top.

I'm not sure about r^{2} value--usually a statistical measure of deviation of data from a model. No data was provided here.???