a Use the projectile model, h= -16t^2+vt+h_0 and substitute the appropriate numbers for v and h_0.
B
b Use the equation from Part A. Let h=0 and solve for t.
A
a h=-16t^2 + 29t + 6
B
b 2 sec
Practice makes perfect
a When we determine an equation that models the height of the shot put, we need to start with the general form for projectiles.
h= -16t^2+ vt+ h_0
We can replace v with the velocity of 29 ft/sec and h_0 with the initial height of 6 ft.
h= -16t^2+ 29t+ 6
b From Part A, we developed the equation that relates height and time for this projectile.
-16t^2+29t+6=h
Since the ground is at zero feet, we can let h=0 and solve for t. To do this, we also need to identify the values of a, b, and c.
ax^2+bx+c=0
⇓
-16t^2+29t+6=0
⇓
a=-16, b=29, c=6
Let's multiply a and c to get -96 and look for factors of -96 that add to 29. Since the product ac is negative in this case, one factor will be positive and one negative. Since b is positive, the factor with the larger magnitude will be positive.
Factors of -96
Sum of factors
-1, 96
95
-2, 48
46
-3, 32
29
Since we arrived at a sum of 29, we do not need to list the rest of the factors. Now, we can split the middle term 29t into -3t + 32t. Let's replace that in our quadratic and then factor.
