An object is thrown vertically upward such
that it has a speed of 32 m/s when it reaches two thirds of its maximum height above the launch point.
Find the maximum height h. The accelera-
tion of gravity is 9.8 m/s2 .
Answer in units of m
solution
To find the maximum height (h) of the object, we can use the given information about its speed at two-thirds of the maximum height.
Let's break down the problem:
1. The initial velocity (u) of the object is given as 32 m/s.
2. The acceleration (a) due to gravity is given as -9.8 m/s² (negative because it acts in the opposite direction of the object's motion).
3. We need to find the maximum height (h) reached by the object.
To solve this problem, we can use the kinematic equation that relates displacement (h), initial velocity (u), final velocity (v), and acceleration (a):
v² = u² + 2aΔh
Since the object is thrown vertically upward and comes to rest at its maximum height, the final velocity (v) is zero.
Therefore, we can simplify the equation:
0 = (32 m/s)² + 2(-9.8 m/s²)h
Simplifying further:
0 = 1024 m²/s² - 19.6 m/s² h
Now, isolate h:
19.6 m/s² h = 1024 m²/s²
Divide both sides of the equation by 19.6 m/s²:
h = 1024 m²/s² / 19.6 m/s²
Calculating this:
h ≈ 52.245 m
Therefore, the maximum height (h) reached by the object is approximately 52.245 meters.