Suppose a diver jumps from a ledge that is 115 feet

Question

Suppose a diver jumps from a ledge that is 115 feet above the ocean and the initial upward velocity is 8 feet per second. The vertical motion of the diver can be modeled by the function h = −t² + 8t + 115. How long will it take until the diver enters the water? How do you know?

Answer 1

The diver enters the water when their height is equal to 0. So what we do is see when their height function is equal to 0.

−t² + 8t + 115 = 0.

We know how to solve this using the quadratic formula! Taking the positive solution, we find t ≈ 15.446.

Hope this helps!

Answer 2

h= -t^2 +8t + 115

set h=0 and solve for t

t^2 -8t -115 = 0

t= 8/2 + or - (1/2)sqr(64+4(115))

t = 4 + .5sqr(524)

t =4 + about 11.45

t = 15.45 seconds to hit the ground

that's if you follow the math

but the t^2 term should be -16t^2 if it's close to physics and the real deceleration due to gravity

using -16t^2 would change the result to

0 =-16t^2 + 8t + 115

16t^2 -8t -115 = 0

t = 8/32 + (1/32)sqr(64+64(115))

t = 1/4 + (1/32)86.16

t = about 2.94 seconds which looks far more realistic as an answer