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distance traveled by a uniformly accelerated bike is given as
here we know that
now we will have from above equation
so it will cover the total distance of 300 m
Answer:
1. The final velocity of the penny before it hits the ground is approximately 54.25 m/s
2. The velocity after falling 45 meters is approximately 37.10 m/s
3. The height up the hill one can start without going over the smaller hill is approximately 2.75 meters
Explanation:
The height of the Bellagio, h = 150 meters
The mass of the penny, m = 1 kg
The kinematic equation of motion that can be used to find the final velocity of the penny 'v' before it hits the ground, is presented as follows;
v² = u² + 2·g·h
Where;
v = The final velocity of the penny after dropping through a height, 'h'
u = The initial velocity of the penny = 0 m/s for the penny initially at rest
g = The acceleration due to gravity ≈ 9.81 m/s²
h = The height from which the penny was dropped = 150 m
∴ v² ≈ 0² + 2 × 9.81 × 150 = 2,943
v ≈ √2,943 ≈ 54.25
The final velocity of the penny before it hits the ground, v ≈ 54.25 m/s
2. Here, the initial velocity, u = 80 km/h = 80 km/h × 1000 m/km × 1 h/(60 × 60 s) = 200/9 m/s = 22. m/s
The height of supreme scream, = 90 meters
The height at which the velocity is required, h = 45 meters
From v² = u² + 2·g·h, we get;
v² = 22.² + 2 × 9.81 × 45 ≈ 1,376.73
∴ v = √1,376.73 ≈ 37.10
The velocity 'v' after falling 45 meters is, v = 37.10 m/s
3. The height of the smaller hill, h = 5 meters
The running start = 4 m/s = The initial velocity
The velocity required to reach the height, h, of the smaller heal v = √(2·g·h)
∴ v = √(2 × 9.81 m/s² × 5 m) ≈ 9.9 m/s
The height 'h'' up the larger hill that will give a velocity, 'v', at the bottom of the smaller hill of approximately 9.9 m/s with an initial velocity, u = 4 m/s, is given as follows;
v² = u² + 2·g·h'
9.9² = 4² + 2 × 9.81 × h'
∴ h' = 9.9²/(4² + 2 × 9.81) ≈ 2.75
Given that the running start is 40 m/s, the height up the hill one can start without going over the smaller hill, h' ≈ 2.75 meters
1) The grapefruit is in free fall, so it moves by uniformly accelerated motion, with constant acceleration 
. Calling h its height at t=0, the height at time t is given by
We are told thatn when
the grapefruit hits the ground, so h(0.75 s)=0. If we substitute these data into the equation, we can find the initial height h of the grapefruit:
2) The speed of the grapefruit at time t is given by
where
is the initial speed of the grapefruit. Substituting t=0.75 s, we find the speed when the grapefruit hits the ground:
We are told thatn when
2) The speed of the grapefruit at time t is given by
where