What minimum speed must the block have at the base of the 70 m hill to pass over the pit at the far (right-hand) side of that hi
1 answer:
Answer:
initial velocity is v = 4.95 m / s
Explanation:
To solve this exercise we use the projectile launch ratios, when the block leaves the hill its speed is horizontal, let's find the time it takes to fall to the other point.
Initial vertical velocity is zero
y = y₀ + v_{oy} t - ½ g t²
y-y₀ = 0 -1/2 g t²
t =
calculate
t =
t = 2.02 s
with this time we can substitute in the horizontal displacement equation
x = v₀ₓ t
v₀ₓ = x / t
suppose that the distance between the two points is x = 10 m
v₀ₓ = 10 / 2.02
v₀ₓ = 4.95 m / s
initial velocity is v = 4.95 m / s
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Answer:
Both balls hit the ground at the same time
Explanation:
Adam drops the ball from rest, so the ball just "<em>falls</em>" in vertical direction, being gravity its only acceleration, for cinematic movements we use that:
In this case we have that gravity is negative, and as Adam drops the ball,
Bob throws the ball horizontally, so the movement will be a <em>parabola</em>, we can divide into horizontal direction, and vertical direction.
But we only need to analize the vertical movement, in wich again the only acceleration is gravity, and compare it with Adam's ball. Again we have that gravity is negative, and as the initial throw is horizontal,
Finally, we have that
where
both for Adam's vertical drop, and for Bob's vertical component of the parabolic throw.
Now, if we put (the origin of the vertical coordinate), we get for both cases that
where we can clear the value for the time t, of the fall, wich will be the same in both cases.
Hence, both balls hit the ground at the same time.
Answer:
Explanation:
We have an uniformly accelerated motion, with a negative acceleration. Thus, we use the kinematic equations to calculate the distance will it take to bring the car to a stop:
The acceleration can be calculated using Newton's second law:
Recall that the maximum force of friction is defined as . So, replacing this:
Now, we calculate the distance: