Please answer those two questions step by step College Physics Course 1. A tiger
ID: 3279787 • Letter: P
Question
Please answer those two questions step by step College Physics Course
1. A tiger leaps horizontally from a 6.2 m -high rock with a speed of 3.6 m/s .
Part A
How far from the base of the rock will she land?
Express your answer using two significant figures.
2. You buy a plastic dart gun, and being a clever physics student you decide to do a quick calculation to find its maximum horizontal range. You shoot the gun straight up, and it takes 5.2 s for the dart to land back at the barrel.
Part A
What is the maximum horizontal range of your gun?
Express your answer using two significant figures and include the appropriate units.
3. An athlete performing a long jump leaves the ground at a 28.9 angle and lands 7.72 m away.
Part A
What was the takeoff speed?
Express your answer using three significant figures and include the appropriate units.
Part B
If this speed were increased by just 5.0%, how much longer would the jump be?
Express your answer using two significant figures and include the appropriate units.
Thank You
Explanation / Answer
If you assume no air resistance, her 3.6 m/s horizontal velocity will be constant until she lands, but for how long? First find the time to free fall (from rest) 6.2m:
H = Vo * t + 1/2 * g * t^2; with
g=9.8 m/s^2
Vo = 0
solve for t:
t = SQRT (2 * 6.2 / 9.8) = 1.12 sec
So she travels horizontally for 1.12 sec at 3.6 m/s, so she lands:
D = V * t = 3.6 * 1.12 = 4.032 m away
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The horizontal range of an object shouted at angle:
R = (2*V^2*sin2Theta) / g
And the maximum range at 45°:
= 2*(g*t^2/4)
= 2*(9.8*5.2^2/4)
= 132.4 m
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The working formula is
R = (V^2)(sin 2A)/g
where
R = range = maximum horizontal distance = 7.72 meters (given)
V = initial velocity
A = 28.9 degrees
g = acceleration due to gravity = 9.8 m/sec^2 (constant)
Substituting appropriate values,
7.72 = V^2(sin 2*28.9)/9.8
Solving for "V",
V = 8.92 m/sec.
Some dart gun.
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If the above speed were increased by 4%, then V = 9.27 m/sec and the corresponding range will be
R = (9.27)^2(sin 2*28.9)/9.8
R =8.71meters
Hence, the jump will be 9.27 -8.71 = 0.56 m longer.