Imagine that you throw a ball vertically upwards from the rooftop of a building. The ball abandons your hand at a point that has the same height as the railway of the rooftop with an ascending speed of 15m/s, been left in free falling. Coming down the ball crosses the railway, at this point gravity is 9.81 m/s2.
Calculate the displacement and velocity of the ball 1 s and 4 s after been let go.
b)The velocity when the ball is 5 m above the railway.
c) Maximum height reached and the moment in which the ball reaches that height
The acceleration of the ball at its maximum height
a. D = Vo*t + 0.5g*t^2
D = 15*1 - 4.9*1^2 = 10.1 m.
V = Vo + g*t = 15 - 9.8*1 = 5.2 m/s
Tr = -Vo/g = -15/-9.8 = 1.53 s. = Rise time.
h max = -(Vo^2)/2g = -(15^2)/-19.6 = 11.48 m. Above launching point.
Tf = 4 - 1.53 = 2.47 s.
D = 4.9*Tf^2 = 4.9*2.47^2 = 29.9 m After
4 s.
V^2 = Vo^2 + 2g*d = 0 + 19.6*29.9 = 585.93
V = 24.2 m/s.
b. V^2 = Vo^2 + 2g*(11.48-5) = 0 + 127
V = 11.27 m/s.
c. h max = 11.48 m.(See previous calculation).
Tr = 1.53 s.(See previous calculation).
To calculate the displacement and velocity of the ball at different times, we can use the equations of motion for freely falling objects.
First, let's consider the time when the ball was let go:
Initial velocity (u) = 15 m/s (upward)
Acceleration due to gravity (a) = -9.81 m/s^2 (negative because it acts downward)
a) After 1 second:
We can use the equation:
s = ut + (1/2)at^2
Substituting the values:
s = (15 m/s)(1 s) + (1/2)(-9.81 m/s^2)(1 s)^2
s = 15 m/s - 4.905 m/s^2
s = 10.095 meters (displacement)
To find the velocity (v) after 1 second:
We can use the equation:
v = u + at
Substituting the values:
v = 15 m/s + (-9.81 m/s^2)(1 s)
v = 15 m/s - 9.81 m/s^2
v = 5.19 m/s (velocity)
b) When the ball is 5 meters above the railway:
To find the time it takes for the ball to reach this height, we can use the equation:
s = ut + (1/2)at^2
Substituting the values:
5 m = (15 m/s)t + (1/2)(-9.81 m/s^2)t^2
This equation is a quadratic equation in terms of time (t). Solving this equation will give us the value of time (t) when the ball is 5 meters above the railway.
c) Maximum height reached and the moment in which the ball reaches that height:
To find the maximum height, we need to consider the point where the velocity of the ball becomes zero (v = 0 m/s).
Using the equation:
v = u + at
Substituting the values:
0 m/s = 15 m/s - 9.81 m/s^2(t)
Solving this equation will give us the value of time (t) when the ball reaches its maximum height.
Once we have the time (t), we can substitute it into the equation:
s = ut + (1/2)at^2
Substituting the values:
s = (15 m/s)(t) + (1/2)(-9.81 m/s^2)(t)^2
This equation will give us the maximum height reached by the ball.
The acceleration of the ball at its maximum height will be equal to the acceleration due to gravity (9.81 m/s^2) since it is the only force acting on the ball at that point.
Please note that to obtain accurate results, it is important to use consistent units throughout the calculations.