i have this problem, it asks:
(a) With what speed must a ball be thrown vertically from ground level to rise to a maximum height of 50m?
answer: 31m/s
(b)How long will it be in the air?\
answer: 6.4s
(c) Sketch graphs of y, v, and a vs. t for the ball. On the first two graphs, indicate the time at which 50m is reached.
i already solved for a and b but i cannot figure how the graph for c would look.
any help is appreciated!
A more accurate answer for the required velocity is sqrt(2gH) = 31.3 m/s
The time in the air is 2*31.3/9 = 6.39 s
y = 31.3 t -4.9 t^2 is height vs. time.
It is an upside down parabola with a maximum at t = 3.2 s
v = 31.3 - 9.8 t is velocity vs time
It is a straight line that goes through zero at t = 3.2 s
a = -9.8 is acceleration vs time
We can't draw graphs for you here.
To sketch the graphs of y (position), v (velocity), and a (acceleration) as functions of time (t) for the ball, we need to consider the motion of the ball throughout its trajectory.
Since the ball is thrown vertically upwards, it experiences an initial positive velocity and negative acceleration due to gravity. As it reaches its maximum height, its velocity becomes zero, and then it starts to come back down with a negative velocity and positive acceleration until it reaches the ground again.
Here's how you can sketch the graphs:
1. Position-Time Graph (y vs. t):
- Let's assume the ball is thrown at time t = 0.
- Initially, the position is at ground level (y = 0).
- As time passes, the ball rises and reaches a maximum height of 50m at some time (let's say t = t₁).
- After reaching the maximum height, the ball descends and reaches the ground again at some time (let's say t = t₂).
Therefore, the position-time graph would look like this:
^
y (m) |
|
50 |-------------- t₁ --------------|------------------------- t₂ -------------------
|
|
|_________________________________________ t (s)
2. Velocity-Time Graph (v vs. t):
- Since the ball is thrown upwards, it initially has a positive velocity.
- As it reaches the maximum height, its velocity decreases until it becomes zero.
- After that, it starts falling downwards, gaining negative velocity.
Therefore, the velocity-time graph would look like this:
^
v (m/s) |
|
+----| |----- 0 ----|------------------------- - v₀ ------------------
|
|
|_________________________________________ t (s)
3. Acceleration-Time Graph (a vs. t):
- The acceleration of the ball is constant and equal to the acceleration due to gravity (g), which is -9.8 m/s² (negative since it's pointed downwards).
- So, the acceleration-time graph would be a horizontal line with a value of -9.8 m/s².
Therefore, the acceleration-time graph would look like this:
^
a (m/s²) |
|
-9.8 |____________________________ t (s)
To indicate the time at which the ball reaches a height of 50m, draw a vertical dotted line from the y-axis (position) graph and show the corresponding points on the velocity and acceleration graphs.