You need to divide the motion into its component: vertical and horizontal motion.
The time taken to fall vertically from the cliff is equal to the time taken to move horizontally.
Using the vertical component, which is an accelerated motion with an initial velocity equal to zero, we can solve for t:
h = 1/2 · g · t²
t = √(2·h / g)
= √(2·50 / 9.8)
= 3.2 s
Horizontally, it is a constant motion:
d = v · t
= 20 · 3.2
= 64 m
The ball will strike the ground at a distance of 64 meters from the cliff.
Ek = (m*V^2) / 2 where m is mass and V is speed, then we can take this equation and manipulate it a little to isolate the speed.
Ek = mv^2 / 2 — multiply both sides by 2
2Ek = mv^2 — divide both sides by m
2Ek / m = V^2 — switch sides
V^2 = 2Ek / m — plug in values
V^2 = 2*30J / 34kg
V^2 = 60J/34kg
V^2 = 1.76 m/s — sqrt of both sides
V = sqrt(1.76)
V = 1.32m/s (roughly)
Answer:
A light bulb lightning up
Explanation:
Near the Earth's surface, a freely falling body has constant acceleration at every instant of its fall ... 9.8 meters per second^2.
Answer:
2.846m
Explanation:
The diver is performing projectile motion.
To find x(final), we are going to use the equation x(final) = v(initial)*t + x(initial)
x(initial) = 0
x(final) = ?
v(initial) = 2.3 m/s
we don't know t
To find t we will use y(final) = 1/2*(-9.8)*t^2 + v(initial in the y dir.)*t + y(initial)
- 9.8 in the acceleration in the y dir.
y(final) = 0
y(initial) = 7.5
v(initial in the y dir.) = 0
If we solve for t we get: t = 1.237s
Now we have all the components to solve for x(final) in x(final) = v(initial)*t + x(initial)
x(final) = 2.3*1.237 + 0
x(final) = 2.846m
