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?
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.
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
above the ground and 7.4 meters above you.
To find out how high the egg will reach when thrown upward, we can use the equations of motion.
The equation that relates the final velocity (v), initial velocity (u), acceleration (a), and distance (s) is:
v^2 = u^2 + 2as
Since we are throwing the egg upward, the initial velocity (u) would be positive (+11 m/s), and the final velocity (v) would be zero (as it will momentarily stop at the highest point). The acceleration (a) can be taken as the acceleration due to gravity, which is approximately -9.8 m/s^2 (taking downward direction as negative).
Substituting the values into the equation, we have:
(0)^2 = (11)^2 + 2(-9.8)s
Simplifying the equation, we get:
0 = 121 - 19.6s
Rearranging the equation to solve for s (height), we have:
19.6s = 121
Dividing both sides by 19.6, we find:
s = 121 / 19.6
Calculating the value, we get:
s ≈ 6.17 meters
Therefore, the egg will reach a height of approximately 6.17 meters above the ground when thrown with a speed of 11 m/s upward.