A ball is projected horizontally from the edge
of a table that is 1.49 m high, and it strikes
the floor at a point 1.01 m from the base of
the table.
The acceleration of gravity is 9.8 m/s
2
.
What is the initial speed of the ball?
Answer in units of m/
To find the initial speed of the ball, we can use the kinematic equation:
h = ut + (1/2)gt^2
Where:
h = height of the table
u = initial velocity (which is what we're trying to find)
g = acceleration due to gravity
t = time of flight
Since the ball is projected horizontally, the time of flight, t, can be determined using the equation:
t = sqrt((2h)/g)
Substituting the given values:
h = 1.49 m
g = 9.8 m/s^2
t = sqrt((2 * 1.49) / 9.8)
t = sqrt(0.303)
Now, we can use the equation h = ut to find the initial velocity, u.
1.01 = u * sqrt(0.303)
Solving for u:
u = 1.01 / sqrt(0.303)
Calculating:
u ≈ 1.85 m/s
Therefore, the initial speed of the ball is approximately 1.85 m/s.
To find the initial speed of the ball, we can use the formula for the horizontal distance traveled by a projectile:
d = V_x * t
Where:
d = horizontal distance traveled (1.01 m)
V_x = initial horizontal velocity (which we need to find)
t = time of flight
First, we need to find the time of flight. We can use the formula for the vertical displacement of a projectile:
h = (1/2) * g * t^2
Where:
h = vertical displacement (1.49 m)
g = acceleration due to gravity (9.8 m/s^2)
t = time of flight
Rearranging the equation, we get:
t = sqrt((2 * h) / g)
Plugging in the given values, we find:
t = sqrt((2 * 1.49) / 9.8) ≈ 0.56 s
Now that we know the time of flight, we can plug it into the first equation to find the initial horizontal velocity:
1.01 = V_x * 0.56
V_x = 1.01 / 0.56 ≈ 1.803 m/s
Therefore, the initial speed of the ball is approximately 1.803 m/s.