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 am not sure about b!! And for c I know I plug 0 in somewhere but I'm not sure where. All help is greatly appreciated, thank you!

Answer 1

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