The longest cave system in the world is the
2 answers:
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<u>Answer:</u>
a) Minimum speed must he drive off the horizontal ramp = 39.78 m/s
b) Minimum speed must he drive off the horizontal ramp with 7° above the horizontal = 23.93 m/s
<u>Explanation:</u>
a) The height of ramp = 1.5 meter
Horizontal distance he must clear = 22 meter
The car is having horizontal motion and vertical motion. In case of vertical motion the acceleration on the car is acceleration due to gravity.
We have equation of motion, , s is the displacement, u is the initial velocity, a is the acceleration and t is the time.
In case of vertical motion initial velocity = 0 m/s, acceleration = 9.8 , we need to calculate time when displacement = 1.5 meter.
So the car has to cover a distance of 22 meter in 2.119 seconds.
So minimum speed required = 22/0.553 = 39.78 m/s
Minimum speed must he drive off the horizontal ramp = 39.78 m/s
b) When the take of angle is 7⁰ the vertical speed of car is not zero = V sin 7 = 0.122 V
So the in case of vertical motion we have initial velocity = 0.122 V, S = -1.5 meter( below ramp), acceleration = -9.8
Substituting
In case of horizontal motion
Horizontal speed of car = V cos 7 = 0.993V
So it has to travel 22 meter in t seconds
0.993Vt = 22, Vt = 22.155 m
Substituting in the equation
We will get
Speed required = 22.155/0.926 = 23.93 m/s
Minimum speed must he drive off the horizontal ramp with 7° above the horizontal = 23.93 m/s
Answer:
h = v₀ g / a
Explanation:
We can solve this problem using the kinematic equations. As they indicate that the air does not influence the vertical movement, we can find the time it takes for the body to reach the floor
y = t - ½ g t²
The vertical start speed is zero
t² = 2t / g
The horizontal document has an acceleration, with direction opposite to the speed therefore it is negative, the expression is
x = v₀ₓ t - ½ a t²
Indicates that it reaches the same exit point x = 0
v₀ₓ t = ½ a t2
v₀ₓ = ½ a (2h / g)
v₀ₓ = v₀
h = v₀ g / a