A child throws a small ball vertically upwards. The ball is caught at the initial height 1.8 seconds after being thrown. What was the net displacement for the ball? What was the initial speed of the ball? What was the final speed of the ball? How high did it rise?
The net displacement was ZERO
That is a trick question
It ended up exactly where it started. It moved a distance, but did NOT displace.
Is the initial speed of the ball 0 m/s and the final speed 0 m/s?
t = .9 seconds up and .9 seconds down
now it is just like the other problem
0 = Vi - 9.81 t
0 = Vi - 9.81(.9)
Vi = 9.81*.9 which is about 9 m/s up
LOL, final speed down is the same as initial speed up also Vi
Note it asks for speed, not velocity, no plus and minus, just speed
h = 1.8 + Vi t - 4.9 t^2
So, can speed be negative, or is velocity only negative?
In your car you measure speed with your speedometer, no negative about it.
Tho get velocity you need both speedometer and compass, speed which is absolute value and also direction. If east is positive, west is negative.
To answer these questions, we need to use the equations of motion for vertical motion under constant acceleration.
1. Net Displacement (change in position):
The net displacement is the difference between the final position and the initial position. In this case, the final position is the same as the initial position since the ball is caught at its initial height. Therefore, the net displacement is 0.
2. Initial Speed (velocity):
The initial speed of the ball can be determined using the equation:
v = u + at
where v is the final velocity (which is zero because the ball is caught), u is the initial velocity (which we need to find), a is the acceleration due to gravity (approximately 9.8 m/s^2), and t is the time.
Since we know that v = 0, and t = 1.8 seconds, we can rearrange the equation to solve for u:
0 = u - (9.8)(1.8)
u = 9.8(1.8)
u = 17.64 m/s
Therefore, the initial speed of the ball is 17.64 m/s.
3. Final Speed:
The final speed of the ball can also be determined using the equation:
v = u + at
Since the ball is caught at its highest point, the final velocity (v) is zero. So we have:
0 = 17.64 - (9.8)(t)
t = 17.64/9.8
t = 1.8 seconds
Therefore, the final speed of the ball is also zero since it comes to a stop when it reaches its highest point.
4. Height Reached:
To find the height reached by the ball, we can use the equation:
s = ut + 1/2at^2
Since we know that u = 17.64 m/s, t = 1.8 seconds, and a = -9.8 m/s^2 (negative because acceleration due to gravity is downwards), we can substitute these values into the equation:
s = (17.64)(1.8) + 0.5(-9.8)(1.8)^2
s = 31.752 - 15.876
s = 15.876 m
Therefore, the ball reaches a height of 15.876 meters above its initial position.