A stone is catapulted vertically upward with an initial velocity of 20m/s what maximum range does it cover
To determine the maximum range covered by the stone, we need to find the height at which the velocity becomes zero (at the highest point of its trajectory). At this point, the stone will start to fall back down.
Using the equation of motion for vertical projectile motion:
v_f = v_i - g*t,
where:
v_f = final velocity (0 m/s at the highest point)
v_i = initial velocity (20 m/s)
g = acceleration due to gravity (-9.8 m/s^2)
t = time
Rearranging the equation,
0 = v_i - g*t,
g*t = v_i,
t = v_i / g.
Substituting the given values,
t = 20 / (-9.8),
t ≈ -2.04 s.
Since time cannot be negative in this context, we'll take the absolute value:
t = 2.04 s.
The total time for the stone to reach the highest point and come back down is twice the time taken to reach the highest point:
total time = 2 * t,
total time = 2 * 2.04,
total time ≈ 4.08 s.
To find the maximum range, we multiply the total time by the horizontal component of the initial velocity:
range = total time * v_i,
range = 4.08 * 20,
range ≈ 81.6 m.
Therefore, the maximum range covered by the stone is approximately 81.6 meters.
To find the maximum range covered by the stone, we need to first determine how long it takes for the stone to reach the top of its trajectory.
The initial vertical velocity of the stone is 20 m/s. Considering the downwards direction as negative, the velocity decreases by 9.8 m/s due to gravity every second. At the top of its trajectory, the vertical velocity becomes zero before the stone starts to fall downwards.
Using the equation of motion for vertical motion:
vf = vi + at
Where:
vf = final velocity (0 m/s at the top)
vi = initial velocity (20 m/s)
a = acceleration (-9.8 m/s^2)
t = time
0 = 20 - 9.8t
Simplifying the equation:
9.8t = 20
t = 20 / 9.8
t ≈ 2.04 seconds
Next, we can find the maximum height reached by the stone using the equation:
vf^2 = vi^2 + 2as
Where:
vf = final velocity (0 m/s)
vi = initial velocity (20 m/s)
a = acceleration due to gravity (-9.8 m/s^2)
s = displacement (maximum height)
0 = (20^2) + 2(-9.8)s
400 = -19.6s
s ≈ -400 / -19.6
s ≈ 20.41 meters
Now, we can calculate the maximum range covered by the stone. Since the motion is symmetrical, the time taken to reach the top is half of the total time of flight (time taken for the stone to ascend and descend). So, the total time of flight is approximately 2.04 seconds * 2 = 4.08 seconds.
Using the equation:
range = horizontal velocity * time
The horizontal velocity remains constant throughout the motion (assuming negligible air resistance). Therefore, we need to find the horizontal component of the initial velocity.
The horizontal component of the initial velocity is given by:
vx = vi * cos(θ)
Where θ is the angle of projection. In this case, the stone is projected vertically, so the angle of projection is 90 degrees. The cosine of 90 degrees is 0.
vx = 20 * 0
vx = 0 m/s
Since the horizontal velocity is zero, the range covered by the stone is also zero.
Therefore, the maximum range covered by the stone is 0 meters.