Question

A projectile launched vertically from an altitude of h0​ feet with an initial velocity of v0​ feet per second. The height of the projectile is h(t) feet after t seconds is given by h(t)=−16t2+v0​t+h0​ If h(1)=143 and h(2)=179, find the values of v0​ and h0​. v0​= h0​=

Answer 1

To find the values of v0 and h0, we can use the given information: h(1) = 143 and h(2) = 179. By substituting these values into the equation h(t) = -16t^2 + v0t + h0, we can find two equations. By solving these equations simultaneously, we get v0 = 76 and h0 = 83.

Step-by-step explanation:

To find the values of v0 and h0, we can use the given information:

We are given that h(1) = 143 and h(2) = 179.

Using the equation h(t) = -16t^2 + v0t + h0, we can substitute the values of t = 1 and t = 2 into the equation:

h(1) = -16(1)^2 + v0(1) + h0 = 143

h(2) = -16(2)^2 + v0(2) + h0 = 179

Simplifying, we get two equations:

-16 + v0 + h0 = 143

-64 + 2v0 + h0 = 179

By subtracting the second equation from the first equation, we can eliminate the h0 term and solve for v0:

48 + v0 - (-64 + 2v0) = 143 - 179

-v0 + 112 = -36

v0 = 112 - 36 = 76

Substituting the value of v0 = 76 into any of the original equations, we can solve for h0:

-16 + 76 + h0 = 143

60 + h0 = 143

h0 = 143 - 60 = 83

Therefore, v0 = 76 and h0 = 83.