A ball is thrown from the top of a building with an initial velocity of 30m/s straight upward, at an initial height of 50m above the ground. The ball just misses the edge of the roof on is way down and hits the ground. Determine: a) time needed for ball to reach max height. B)the max height. C) time needed for ball to hit ground. D) velocity and diplacement ( magnitude and direction) of the ball at t=sec( measured from beginning. Please explain step by step

What did you use to find the time for the egg to reach it's maximum height? The answer that you put is kinda confusing.

v = Vi - g h

v is 0 at top
so 0 = Vi - 9.81 t
or t = Vi/9.81 This should be embedded in your soul

You need it for every problem like this

remember, vertical velocity is zero at the top

and horizontal velocity is constant the whole time

I don't understand the maximum height part? what is the formula?

Hey, you throw something straight up at velocity +Vi

The force on it is down, -m g
F = m a
-m g = m a
so
a = -g
so
v(t) = -g t + a constant
since v(0) = Vi the constant is Vi
v(t) = Vi - g t
NOW when it gets to the top, the velocity v( t at top) = 0
so at top
0 = Vi - g t
SO
t = Vi/g AT THE TOP

then for height
h = Hi + Vi t - 4.9 t^2

To determine the answers, we can break down the problem into different parts using the concepts of projectile motion and basic kinematics equations. Let's start step by step:

a) Time needed for the ball to reach its maximum height:
1. The initial velocity is given as 30 m/s straight upward.
2. The final velocity at the maximum height will be zero since the ball momentarily stops before falling down.
3. The acceleration due to gravity is always acting downward and has a value of approximately 9.8 m/s² (neglecting air resistance).
4. Use the kinematic equation: vf = vi + at, where vf is the final velocity, vi is the initial velocity, a is the acceleration, and t is the time.
Plug in the given values: 0 m/s = 30 m/s - 9.8 m/s² × t.
5. Rearrange the equation to solve for time: t = (30 m/s) / 9.8 m/s².

b) The maximum height reached by the ball:
1. The maximum height occurs when the velocity of the ball is zero.
2. Use the kinematic equation: vf² = vi² + 2ad, where d is the vertical displacement.
Plug in the given values: 0 m/s = (30 m/s)² - 2 × (-9.8 m/s²) × d.
3. Rearrange the equation to solve for vertical displacement: d = (30 m/s)² / (2 × 9.8 m/s²).

c) Time needed for the ball to hit the ground:
1. The initial velocity after reaching maximum height will be the same magnitude as the initial velocity, but in the opposite direction.
2. The vertical displacement from maximum height to the ground is equal to the initial height (50 m).
3. Use the kinematic equation: vf² = vi² + 2ad.
Plug in the values: 0 m/s = (-30 m/s)² + 2 × (-9.8 m/s²) × (-50 m).

d) Velocity and displacement of the ball at t = 1 second:
1. Use the kinematic equation: d = vit + 0.5at², where d is the displacement, vi is the initial velocity, a is the acceleration, and t is the time.
Plug in the values: d = (30 m/s) × 1 s + 0.5 × (-9.8 m/s²) × (1 s)².
2. The displacement will be negative since the ball is moving downward, while the initial velocity is positive (upward).
3. To find the velocity, use the kinematic equation: vf = vi + at.
Plug in the values: vf = 30 m/s - 9.8 m/s² × 1 s.

By solving these steps, you will get the answers to each part of the problem.

Vi = 30

v = Vi - 9.81 t
at top v = 0
so t at top = 30 / 9.81

50 + 30 t - 4.9 t^2

t plus time needed to fall from part B height

0 = 30 + Vi(total time) - 4.9 total time^2
solve for total time

v = Vi - gt the whole time
h = 30 + Vi t - 4.9 t^2