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:
The maximum height the box will reach is 1.72 m
Explanation:
F = k·x
Where
F = Force of the spring
k = The spring constant = 300 N/m
x = Spring compression or stretch = 0.15 m
Therefore the force, F of the spring = 300 N/m×0.15 m = 45 N
Mass of box = 0.2 kg
Work, W, done by the spring = 
and the kinetic energy gained by the box is given by KE = 
Since work done by the spring = kinetic energy gained by the box we have
= therefore we have v = 
= 
= 
= 5.81 m/s
Therefore the maximum height is given by
v² = 2·g·h or h = 
= 
= 1.72 m
deceleration or rėtardation i’m pretty sure (it won’t let me say the second word but it’s correct)
