A ball is projected horizontally from the edge
of a table that is 1.2 m high, and it strikes the
floor at a point 1.11 m from the base of the
table.
The acceleration of gravity is 9.8 m/s2 .
What is the initial speed of the ball?
Answer in units of m/s
Using the height H of the table, compute the time t required to hit the ground.
t = sqrt(2H/g)= 0.495 s
Initial speed =
= horizontal V component = (1.11 m)/t
To find the initial speed of the ball, we can use the kinematic equation that relates the horizontal distance, vertical distance, and initial speed of an object in projectile motion. The equation is:
d = v₀t + (1/2)at²
Where:
- d is the horizontal distance (1.11 m),
- v₀ is the initial speed (what we are trying to find),
- t is the time of flight (what we will calculate), and
- a is the acceleration due to gravity (-9.8 m/s² vertically downward).
Since the ball is projected horizontally, there is no initial vertical velocity, so we can ignore the initial vertical displacement. This means the vertical distance, d, is the height of the table, which is 1.2 m. So, our equation becomes:
1.2 m = 0 + (1/2)(-9.8 m/s²)t²
Simplifying, we have:
1.2 m = (-4.9 m/s²)t²
To solve for t, we can rearrange the equation:
t² = (1.2 m) / (-4.9 m/s²)
t² = -0.2449 s²
Since time cannot be negative, we take the positive square root:
t = √(-0.2449 s²)
t ≈ 0.4949 s
Now that we have the time of flight, we can substitute this value back into the horizontal equation:
1.11 m = v₀ (0.4949 s)
Solving for v₀:
v₀ = 1.11 m / 0.4949 s
v₀ ≈ 2.2408 m/s
Therefore, the initial speed of the ball is approximately 2.2408 m/s.