First, calculate how long the ball is in midair. This will depend only on the vertical displacement; once the ball hits the ground, projectile motion is over. Since the ball is thrown horizontally, it originally has no vertical speed.
t = time vi = initial vertical speed = 0m/s g = gravity = -9.8m/s^2 y = vertical displacement = -45m
y = .5gt^2 [Basically, in this equation we see how long it takes the ball to fall 45m] -45m = .5 (-9.8m/s^2) * t^2 t = 3.03 s
Now we know that the ball is midair for 3.03s. Since horizontal speed is constant we can simply use:
x = horizontal displacement v = horizontal speed = 25m/s t = time = 3.03s
x = v*t x = 25m/s * 3.03s = 75.76 m Thus, the ball goes about 75 or 76 m from the base of the cliff.
I think its B or D, most likely D.
Answer:
4.535 N.m
Explanation:
To solve this question, we're going to use the formula for moment of inertia
I = mL²/12
Where
I = moment of inertia
m = mass of the ladder, 7.98 kg
L = length of the ladder, 4.15 m
On solving we have
I = 7.98 * (4.15)² / 12
I = (7.98 * 17.2225) / 12
I = 137.44 / 12
I = 11.45 kg·m²
That is the moment of inertia about the center.
Using this moment of inertia, we multiply it by the angular acceleration to get the needed torque. So that
τ = 11.453 kg·m² * 0.395 rad/s²
τ = 4.535 N·m
(2)<span>less than 750 N.( if the downward acceleration of elevator were g,then answer would be 0 N.)</span>
