After scoring the championship-winning touchdown...

Question

After scoring the championship-winning touchdown, a football player throws the ball directly upward from the ground with an initial velocity of 52ft/s. Use the formula for a projectile ℎ = −16t^2+vot+ho where t is in seconds, vo is in ft/s and ℎo is in feet. (the o's are dropped down some between the v&t and h as well)

a. When will the ball strike the ground?

b. When will the ball be 40 ft above the ground?

Answer 1

The initial velocity is 52ft/s, so v₀ = 52. The ball is thrown straight from the ground, so the initial height is 0ft, so h₀ = 0.

h = −16t²+v_ot+h_o = −16t² + 52t + 0

h = −16t² + 52t

The ball will strike the ground when the height is 0 again. Substitute 0 for h, then factor to solve the equation.

0 = −16t² + 52t

0 = t(-16t + 52)

either t = 0, or -16t + 52 = 0. t = 0 is when the ball was thrown, not when it hits the ground, so solve the second equation.

-16t + 52 = 0

-16t = -52

t = 3.25

The ball will hit the ground after 3.25 seconds.

The ball will be 40 ft above the ground when h = 40. Substitute 40 for h, and subtract from both sides to write in standard form. Factor to solve the equation.

40 = −16t² + 52t

0 = −16t² + 52t - 40

0 = −4(4t² + 13t + 10)

0 = -4(4t - 5)(t - 2)

4t - 5 = 0 and t - 2 = 0

t = 5/4 and 2

The ball will be 40 ft above the ground at 5/4 = 1.25 seconds and at 2 seconds.

Answer 2

h(t) = -16t^2 + vot + ho. vo = initial velocity, ho = intiial height

h(t) =-16t^2 +52t + 6. assume he's about 6 feet tall

take the derivative and set equal to zero to find time at maximum height

h'(t) = -32t +52 = 0

t = 52/32 = 13/8 = 1 5/8 second to reach = 1.625 sec.

plug that into the original equation to solve for max height

h(13/8) =-16(13/8)^2 + 52(13/8) + 6 = 48.25 feet = max height

to reach 40 feet, t = a little less or more than 1.625 seconds, when it reaching 40 on the way up and then again on the way down.

for h=40, 40=-16t^2 +52t + 6

16t^2 -52t +34 = 0

8t^2 -26t + 17 = 0

t = 26/16 + or - (1/16)sqr(26^2 -4(8)(17)) = about 0.91 seconds or 2.34 seconds

both equidistant in time from the time of maximum height

h=0 is when it hits the ground

h(t) = 0 = -16t^2 +52t + 6

16t^2 -52t -6 = 0

8t^2 -26t -3 = 0

use the quadratic formula

t= 26/16 + (1/16)sqr(26^2 + 4(8)(3)) = about 3.36 seconds

change the solutions if you use ho as a number other than 6 feet