An object is launched directly upward at an initial speed of 42.3 m/s. Assuming there is no air-resistance acting on the object, what is its maximum height relative to its launch position?
KE=PE
KE = m•v²/2
PE = m•g•h
h = v²/2•g
To find the maximum height reached by the object, we need to utilize the laws of motion and kinematics. We can break the problem down into smaller steps:
Step 1: Determine the initial velocity (given):
The initial velocity is provided in the question. It is given that the object is launched with an initial speed of 42.3 m/s.
Step 2: Calculate the time taken to reach the maximum height:
Since the object is launched directly upward and there is no air resistance, it will experience a constant deceleration due to gravity. This deceleration is equal to -9.8 m/s^2 (taking into account the direction).
To find the time taken to reach the maximum height, we can use the kinematic equation for vertical motion:
v = u + at,
where:
v = final velocity (0 m/s at the maximum height),
u = initial velocity (42.3 m/s),
a = acceleration (-9.8 m/s^2),
t = time taken.
Rearranging the equation and solving for t:
0 = 42.3 - 9.8t
9.8t = 42.3
t = 42.3 / 9.8
t ≈ 4.32 seconds.
Thus, it will take approximately 4.32 seconds for the object to reach its maximum height.
Step 3: Calculate the maximum height:
To find the maximum height, we can use another kinematic equation for vertical motion:
s = ut + (1/2)at^2,
where:
s = displacement or height,
u = initial velocity (42.3 m/s),
t = time taken (4.32 seconds),
a = acceleration (-9.8 m/s^2).
Substituting the given values:
s = 42.3 * 4.32 + (1/2) * (-9.8) * (4.32)^2
s ≈ 91.6 meters.
Therefore, the maximum height reached by the object relative to its launch position is approximately 91.6 meters.