Answer:
sin (θ) = y / h
sin (17.3) = y / (10.5)
y = 3.1224 m/s
to get the horizontal (x-component) the equation is as follows:
cos (θ) = x / h
cos (17.3) = x / (10.5)
x = 10.02499 m/s
Using the kinematics equations we can find the answers:
1) v(velocity) = v0 (initial velocity) + A(acceleration) x T(time)
2) v^2 = v0^2 + (2 x A x Δd(change in distance))
3) X(position) = X0(initial position) + (v0 x T) + (1/2 x A x T^2)
to find out how high he jumped use the first to deter mine how long it took to reach his max height (when v = 0), then plug in all the information into equation 3 for the answer:
Equation 1:
v = 0
v0 = 3.1224
a = -9.8
v = v0 + Ax T
( 0) = ( 3.1224) + (-9.8) x(T)
-3.1224 = -9.8T
T = .3186 s
Equation 3:
v = 0
v0 = 3.1224 (vertical direction)
a = -9.8 (vertical direction)
t = 0.3186
X0 = 0
X = X0 + (v0 x T) + (1/2 x A x T^2)
X = (0) + ((3.1224) x (0.3186)) + (1/2 x (-9.8) x (0.3186)^2)
X = 0 + 0.9953 + (-0.49737)
X = 0.49 m <===.ANSWER
to find out how far he jumped is simpler. Since it took the jumper ).3186 sec to reach his peak height it must take double that time to hit the ground again so for this section our new T = 0.6372 s. We can now use the third equation to solve this
Equation 3:
v = 0
v0 = 10.02499 (horizontal direction)
a = 0 (horizontal direction)
t = 0.6372
X0 = 0
X = X0 + (v0 x T) + (1/2 x A x T^2)
X = (0) + ((10.02499) x (0.6372)) + (1/2 x (0) x (0.6372)^2)
X = 0 + 6.3879 + 0
X = 6.39 m <=== ANSWER
