An object is thrown vertically upward such
that it has a speed of 56 m/s when it reaches
two thirds of its maximum height above the
launch point.
Find the maximum height h. The acceleration of gravity is 9.8 m/s
2
.
Answer in units of m
At 2/3 of its maximum height, it has 1/3 of its initial kinetic energy left. That means it has 1/sqrt3 = 0.5773 of its initial velocity. The initial velocity must have been
Vo = 56/0.5773 = 96.99 m/s.
Max height is Vo^2/(2*g)
= 480 m
v.=56m/s
v=0
g=-10m/s2
v2=V.2+2gyx
0=3136-2.9.8.x
19.6x=3136m
x=160m
1/3x=160m
x=3.160=480m
To find the maximum height, we need to use the equations of motion for vertical motion.
Let's assume:
Initial velocity (u) = 56 m/s (upward)
Final velocity (v) = 0 m/s (at the max height, the object momentarily stops and changes direction)
Acceleration (a) = -9.8 m/s^2 (since the object moves against gravity)
Using the equation, v^2 = u^2 + 2as, we can find the displacement (s) or change in height.
0^2 = (56)^2 + 2(-9.8)s
0 = 3136 - 19.6s
19.6s = 3136
s = 3136 / 19.6
s = 160 m
This represents two-thirds of the maximum height, so the maximum height (h) is:
h = (3/2) * s
h = (3/2) * 160
h = 240 m
Therefore, the maximum height (h) is 240 meters.
To find the maximum height of the object, we need to analyze the motion of the object when it reaches two-thirds of its maximum height.
Let's denote the maximum height as h and the initial velocity of the object as v₀.
We know that when the object reaches two-thirds of its maximum height, its velocity is equal to 56 m/s. Using this information, we can set up an equation to calculate v₀.
When the object is at two-thirds of its maximum height, its final velocity, v, is 56 m/s, and its initial velocity, u, is unknown. The acceleration, a, in this case, is the acceleration due to gravity, which is -9.8 m/s² (negative because the object is moving in the opposite direction to gravity).
Using the equation of motion: v² = u² + 2as
We can substitute the values into the equation:
56² = u² + 2(-9.8)h
Simplifying the equation, we get:
3136 = u² - 19.6h
Since the object is thrown vertically upward, its final velocity when it reaches the maximum height is 0 m/s. Therefore, we can set v = 0 and solve for u:
0² = u² + 2(-9.8)h
Simplifying the equation, we get:
0 = u² -19.6h
Now we have two equations:
3136 = u² - 19.6h
0 = u² - 19.6h
We can solve these equations simultaneously to find the values of u and h.
Subtracting the second equation from the first equation, we get:
3136 - 0 = u² - u² - 19.6h + 19.6h
3136 = 0
This equation is not valid, meaning that there is no real solution. Therefore, we made an error in assuming that the object reaches two-thirds of its maximum height with a velocity of 56 m/s.
To find the maximum height h, we need to reevaluate the problem and consider the correct information or clarify the question further.