II.You are standing on the edge of a 50m high cliff and throw a ball vertically up with initial speed of 20m/s: 1.After how long will the ball hit the ground below?
2.Draw the v vs.t graph
Plot the height vs time function
Y (measured from ground below) = 50 + 20 t - 4.9 t^2.
(The 4.9 represents 1/2 the acceleration of gravity, in m/s^2)
Solve for the time when Y = 0. It should be between 5 and 6 seconds.
We do not provide graph-drawing services.
To find out when the ball will hit the ground below, we can use the equations of motion. The key equation we will use for this problem is:
h = ut + (1/2)gt^2
Where:
- h is the height (in this case, 50m)
- u is the initial velocity (in this case, -20m/s, as the ball is thrown upward)
- g is the acceleration due to gravity (which is approximately 9.8m/s^2)
- t is the time we are trying to find
1. The first step is to rearrange the equation to solve for t:
h = ut + (1/2)gt^2
50 = -20t + (1/2)(9.8)t^2
0 = 0.5(9.8)t^2 - 20t + 50
Next, we can solve this quadratic equation using the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / 2a
For this equation, a = 0.5(9.8), b = -20, and c = 50. Plugging these values into the quadratic formula, we get:
t = (-(-20) ± √((-20)^2 - 4(0.5(9.8))(50))) / (2 * 0.5(9.8))
t = (20 ± √(400 - 98(50))) / (9.8)
t = (20 ± √(400 - 490)) / (9.8)
t = (20 ± √(-90)) / (9.8)
Since we cannot take the square root of a negative number when working with real numbers, it means the ball will never hit the ground again. This result makes sense since we threw the ball upward, and it will not return to the ground.
2. However, let's still draw the velocity vs. time graph as requested. To do this, we need to determine the velocity of the ball at any given time.
The velocity of the ball can be found using the equation:
v = u + gt
For this problem, the initial velocity (u) is -20m/s, and the acceleration due to gravity (g) is -9.8m/s^2 (negative because it acts in the opposite direction to the initial velocity).
Now, we can plot the graph. Since we are throwing the ball upwards, the initial velocity is negative, and as the ball continues to rise, the velocity decreases until it reaches zero at the highest point. After that, the velocity becomes negative as the ball starts falling back down.
So the graph would start at -20m/s and decrease until it reaches 0m/s at the highest point, then it becomes negative and increases as the ball falls down.