A steel ball rolls off a flat table of height 1.05 m with a velocity of 1.90 m/s
a) How long will it take the ball after leaving contact with the table to hit the ground?
b) How far from the edge of the table will the ball hit the floor?
To solve this problem, we can use the equations of motion to find the time taken and the horizontal distance traveled by the ball.
Let's start by finding the time taken for the ball to hit the ground.
a) Equation of motion for vertical motion:
h = ut + (1/2)gt^2
Here, h = 1.05 m (height of the table)
u = 0 m/s (initial vertical velocity)
g = 9.8 m/s^2 (acceleration due to gravity)
t = time taken (unknown)
Plugging in the values, we get:
1.05 = 0*t + (1/2)*9.8*t^2
1.05 = 4.9*t^2
Rearranging the equation:
t^2 = 1.05/4.9
t^2 = 0.2143
Taking the square root of both sides:
t = √0.2143
t ≈ 0.463 s
Therefore, it will take approximately 0.463 seconds for the ball to hit the ground.
b) The horizontal distance traveled can be calculated using the equation of motion for horizontal motion:
s = ut
Here, s = horizontal distance traveled (unknown)
u = 1.90 m/s (initial horizontal velocity)
t = 0.463 s (time taken from part a)
Plugging in the values, we get:
s = 1.90 * 0.463
s ≈ 0.8797 m
Therefore, the ball will hit the floor approximately 0.8797 meters from the edge of the table.
To solve these problems, we can use the equations of motion and the principles of kinematics. Let's break down each question and solve them step by step:
a) How long will it take the ball after leaving contact with the table to hit the ground?
To find the time it takes for the ball to hit the ground, we need to use the equation for vertical motion:
d = v_i*t + 1/2 * a * t^2
where:
d = height of the table = 1.05 m
v_i = initial velocity in the vertical direction = 0 m/s (since the ball is rolling off the table)
a = acceleration due to gravity = -9.8 m/s^2 (taking downward direction as negative)
We want to find t, so we can rearrange the equation as follows:
1.05 = 0*t + 1/2 * (-9.8) * t^2
1.05 = -4.9 * t^2
Rearranging further, we get:
t^2 = 1.05 / -4.9
t^2 = -0.2143
Since time cannot be negative, we discard the negative solution. Therefore:
t = √(0.2143)
t ≈ 0.463 s
So, it will take approximately 0.463 seconds for the ball to hit the ground after leaving contact with the table.
b) How far from the edge of the table will the ball hit the floor?
To find the horizontal distance traveled by the ball, we use the equation:
d = v_x * t
where:
d = horizontal distance
v_x = horizontal component of velocity = 1.90 m/s
t = time taken to hit the ground = 0.463 s
Plugging in the values, we have:
d = 1.90 m/s * 0.463 s
d ≈ 0.8797 m
Therefore, the ball will hit the floor approximately 0.8797 meters from the edge of the table.
a. h = 0.5g*t^2.
g = 9.8 m/s^2.
h = 1.05 m.
t = ?
b. Dx = Xo*t.
Xo = 1.90 m/s.
t = Value calculated in part a.