a ball roles off a horizontal table at 0.30 m/s and strikes the ground 0.15m horizontally from the base of the table. (a) how high is the table (b) if another ball roles off the same table at 0.60 m/s how far from the base of the table does it hit.
To find the height of the table, we can use the formula for the horizontal distance covered by a projectile:
1. Let's consider the motion of the ball when it rolls off the table. The initial vertical velocity is zero since it rolls off horizontally.
2. We know that the vertical distance covered by the ball can be calculated using the equation for vertical motion:
h = (1/2) * g * t^2,
where h is the height, g is the acceleration due to gravity (approximately 9.8 m/s^2), and t is the time taken to reach the ground.
3. To determine the time it takes for the ball to reach the ground, we can use the horizontal distance covered by the ball (0.15 m) and the initial horizontal velocity (0.30 m/s):
t = (0.15 m) / (0.30 m/s).
This is because the time taken to reach the ground horizontally is the same as the time taken to reach the ground vertically.
Therefore, t = 0.5 seconds.
4. Now we can substitute the time value into the equation for vertical distance to find the height of the table:
h = (1/2) * (9.8 m/s^2) * (0.5 s)^2.
Solving this equation gives:
h ≈ 1.225 m.
Therefore, the height of the table is approximately 1.225 meters.
To determine the distance from the base of the table where the second ball hits the ground:
1. We can use the same formula for horizontal distance with the given values of horizontal velocity and time:
d = (0.60 m/s) * (0.5 s).
Solving this equation gives:
d = 0.30 m.
Therefore, the second ball will hit the ground approximately 0.30 meters from the base of the table.
H=0.5 GT^2+Vyt
Vx=d/t
g=9.81m/sec
a.
Vx=0.3 d=0.15
0.3=0.15/t
t=2secs
H=0.5(9.81)(2)^2+0
H=19.62m
b.
0.6=d/2
d=1.2m