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.
Answer:
We know from the basic speed distance relation that
Since the car started from rest and it covered the distance between the 2 officer's in 19 minutes we have speed of the car
Which clearly exceeds the limit of 
Answer:
Explanation:
Given:
- mass of solid disk,
- radius of disk,
- force of push applied to disk,
- distance of application of force from the center,
<em>For the condition of no slip the force of static friction must be greater than the applied force so that there is no skidding between the contact surfaces at the contact point.</em>
where:
= static frictional force
Answer:
25032.47 W
Explanation:
Power is the time rate of doing work, hence,
P = Work done(non conservative) / time
Work done (non conservative) is given as:
W = total K. E. + total P. E.
Total K. E. = 0.5mv²- 0.5mu²
Where v (final velocity) = 7.0m/s, u (initial velocity) = 0m/s
Total P. E. = mgh(f) - mgh(i)
Where h(f) (final height) = 7.2m, h(i) (initial height) = 0 m
=> W = 0.5mv² - mgh(f)
P = [0.5mv² - mgh(f)] / t
P = [(0.5*790*7²) - (790*9.8*7.2)] / 3
P = (19355 + 55742.4) / 3 = 75097.4/3
P = 25032.47 W
