Arturo O. answered • 09/15/17
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I assume he jumped up to a maximum height of 1.5 m and then landed at a height of 0.7 m. He had to jump at some initial vertical launch speed v. Did you account for that speed in your calculations?
h(t) = -4.9t2 + vt + 0 = -4.9t2 + vt
You need to find v. You can apply the kinematic relation between the speed at the top, which is zero, and the initial launch speed v.
vf2 - vi2 = 2ahmax
02 - v2 = 2(-9.8)(1.5) m2/s2
v = √29.4 m/s ≅ 5.422 m/s
Now you have the complete expression for h(t):
h(t) = -4.9t2 + 5.422t
To find time to reach h = 0.7,
0.7 = -4.9t2 + 5.422t
-4.9t2 + 5.422t - 0.7 = 0
Solve the quadratic for t and get
t ≅ 0.149 s and 0.957 s
The student reaches 0.7 m on the way up 0.149 seconds after jumping, and again on the way down 0.957 s after jumping.
Arturo O.
I assume you want to find the time when the student lands on the bed. The problem statement says the height of the bed is 0.7 m. You have an equation for the height h as a function of time t. Solve for the t that makes h equal to the height of the bed.
h(t) = 0.7 m
Solve for t. There is no need to solve separately for a t on the way up and a t on the way down, to be added later. Since h(t) is a quadratic, there are 2 solutions. The lower solution is the time it takes to reach 0.7 m on the way up, and the higher solution is the time it takes to reach 0.7 m on the way down, with both times referenced to the moment of the jump, which is t = 0. The answer you seek is the time it takes to land on the bed, which is the higher of the two times.
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09/15/17
Gnarls B.
09/15/17