Projectile Problem

Question

The height, h, in meters, at some time, t, in seconds, of an object propelled upward at a speed of 82 meters per second from an initial height of 103 meters is given by the function h(t)=-10t^2+82t+103

 

a.) What is the maximum height of the object?

b.) When will the object be above 200 meters high?

c.) When will the object hit the ground?

 

I got 410 meters for a. I'm not sure about b and I know for c I plug 0 in but I'm not completely sure. All help is greatly appreciated. Thank you!!

Answer 1

a.)

 

The x-coordinate of the vertex is found by (-82)/((2•(-10)) = 4.1. Maximum height is reached at t = 4.1 seconds.

 

The y-coordinate of the vertex is h(4.1) = -10(4.1)² + 82(4.1) + 103 = -168.1 + 336.2 + 103 = 271.1 meters.

 

b.)

 

h(t) = 200 = -10t² + 82t + 103

 

   0 = -10t² + 82t - 97

 

Using the quadratic formula yields two (2) values for t: passing 200 m on the way up and passing 200 m on the way down.

 

t ≈ 1.43 sec. t ≈ 6.77 sec.

 

c.)

 

When the object hits the ground, h(t) = 0.

 

0 = -10t² + 82t + 103

 

Using the quadratic formula, t ≅ 9.31 sec. at the time the object hits the ground.

 

 

Answer 2

h(t) =-10t² + 82t  + 103  where a= -10     b= 82      c= 103

a) Maximum height means we have to find the vertex, so first find t = -b÷2a

so t = -82÷2*-10 = 4.1, substituting the value of t, we get

h(t) =-10(4.1)² + 82(4.1) + 103 = 271.1, is the maximum height.

 

b.)  200 = -10t² + 82t + 103 

 

-10t² + 82t  -97=0

solve for t by using quadratic formula     t = -b± √ (b²-4ac)/ 2a

 

 

 so t =1.433 and t= 6.77, the object be above 200 meters high at 1.43 seconds while going above and then while coming down at 6.77 seconds.

 

c.) the object will hit the ground when h(t) = 0, so

 

-10t² + 82t + 103 = 0, solve the equation using quadratic formula, we get t = 9.30, the object will hit the ground at 9.30 seconds

Answer 3

a)  To find the maximum height, we use find the vertex (L, k). 

 

L = -b / 2a

 

k = h(L)

 

where:

a = -10

b = 82

c = 103

 

L = -82 / (2*-10)

L = -82 / -20

L = 4.1

 

Then evaluate h(L) to get the maximum height.

 

 

b)

 

Set h(t)=200 and solve for t.

 

200 = -10t² + 82t + 103

 

 

Subtract 200 on both sides of the equation to get a quadratic equation.

 

0 = -10t² + 82t - 97

 

 

Use the quadratic formula:

 

t = (-b ± √(b² - 4ac)) / 2a

 

where:

a = -10

b = 82

c = -97

 

Plug in those values into the formula to get your t values.  Accept the positive value of t.  If both are positive, accept the larger value of t.

 

 

c)

 

Set h(t) equal to 0 and solve for t.

 

0 = -10t² + 82t + 103

 

where:

a = -10

b = 82

c = 103

 

 

Use the quadratic formula from part b to solve for t.