Question

You drop your egg off of a 3 story building (12m above the ground).

You throw it up with speed 11m/s

How high above the ground will the upward thrown egg reach in m?

Answer 1

You are 12 meters high when you throw it up
How high above you will it go?
Vi = 12
v = Vi- g t
0 = 12 - 9.81 t
t = 1.22 seconds upward
H above you = Vi t - 4.9 t^2
= 12 (1.22) - 4.9 (1.22)^2
= 14.7 - 7.3
= 7.4 meters above roof
so it reaches 7.4 + 12 = 19.4 meters

Answer 2

To solve this problem, we can use the equations of motion for a freely falling object and a projectile in free fall. We know that the egg will be in free-fall after it is dropped, so we can use:

h = 0.5*g*t^2

where h is the height above the ground, g is the acceleration due to gravity (9.8 m/s^2), and t is the time in seconds. For the dropped egg, we can set h = 12m and solve for t:

12 = 0.5*9.8*t^2

t^2 = 12/4.9

t = sqrt(2.45)

t ≈ 1.57 s

So it takes 1.57 seconds for the egg to hit the ground after being dropped.

For the upward thrown egg, we can use the equation of motion for a projectile in free fall:

h = v0*t + 0.5*g*t^2

where h is the height above the ground, v0 is the initial upward velocity (11 m/s), g is the acceleration due to gravity (9.8 m/s^2), and t is the time in seconds. To find the maximum height, we can set the velocity at the maximum height to zero:

0 = v0 + g*t_max

t_max = -v0/g

t_max = -11/(-9.8)

t_max ≈ 1.12 s

So the egg reaches its maximum height at 1.12 seconds after being thrown upwards. We can now use this time to find the height:

h = v0*t_max + 0.5*g*t_max^2

h = 11*1.12 + 0.5*9.8*(1.12)^2

h ≈ 17.8 m

Therefore, the upward thrown egg will reach a height of approximately 17.8 meters above the ground.

Answer 3

above the ground and 7.4 meters above you.