A ball thrown straight up climbs for 3.0 sec before falling. Neglecting air resistance, with what velocity was the ball thrown?
1 answer:
Answer:
Speed, u = 29.4 m/s
Explanation:
Given that, A ball thrown straight up climbs for 3.0 sec before falling, t = 3 s
Let u is speed with which the ball is thrown up. When the ball falls, v = 0
Using first equation of motion as :
v = u + at
Here, a = -g
So, u = g × t
u = 29.4 m/s
So, the speed with which the ball was thrown is 29.4 m/s. Hence, this is the required solution.
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The value of the force, F₀, at equilibrium is equal to the horizontal
component of the tension in string 2.
Response:
- The value of F₀ so that string 1 remains vertical is approximately <u>0.377·M·g</u>
Given:
The weight of the rod = The sum of the vertical forces in the strings
Therefore;
M·g = T₂·cos(37°) + T₁
The weight of the rod is at the middle.
Taking moment about point (2) gives;
M·g × L = T₁ × 2·L
Therefore;
Which gives;
F₀ = T₂·sin(37°)
Which gives;
- F₀ ≈ <u>0.377·M·g</u>
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