A ball is thrown at a speed of 39.05m/s at an angle of 39.8degrees above the horizontal. Determine:
V1perpendicular component:
V1Sin(theta)
= 25m/s
V1paralell component:
V1Cos(theta)
= 30m/s
a) the ball's location 2s after being thrown.
V2paralell = V1paralle = 30m/s
V2perpendicular = V1+at
= 25+(-10)(2)
= 5m/s
V2 = sqrt of 30^2 + 5^2
= 30.41m/s
b) the time it takes to reach maximum height.
t = (V2perp - V1perp)/a
= -25/-10
= 2.5s
Note: maximum height will be 0, right? It stops momentarily before it goes down.
c) the maximum height above ground.
V2^2perp = V1^2perp + 2ad
= 25^2 + 2(-10)d
-625 = -20d
-625/-20 = d
d = 31.25m
d) the position and velocity 3s after being thrown.
V2perp = V1perp + at
= 25 + (-10)(3)
= -5m/s
d = v1perp(t) + 1/2at
= (25x3) + .5(-10)(3)
= 60m
To determine the ball's location 2 seconds after being thrown, you'll need to calculate the components of its velocity at that time.
The perpendicular component of the velocity (V2perpendicular) can be found using the equation V2perpendicular = V1perpendicular + at, where V1perpendicular is the initial perpendicular velocity (which we found to be 25m/s) and a is the acceleration due to gravity (-10m/s^2). Plugging in these values, we get:
V2perpendicular = 25m/s + (-10m/s^2)(2s) = 5m/s
The parallel component of the velocity (V2parallel) remains constant throughout the motion and is equal to the initial parallel velocity (V1parallel) which we found to be 30m/s.
To find the total velocity (V2), we can use the Pythagorean theorem:
V2 = √(V2parallel^2 + V2perpendicular^2) = √((30m/s)^2 + (5m/s)^2) = √(900m^2/s^2 + 25m^2/s^2) ≈ 30.41m/s
Therefore, 2 seconds after being thrown, the ball's location can be determined by its total velocity V2, which is approximately 30.41m/s.
To determine the time it takes for the ball to reach its maximum height, you need to find the time when its perpendicular velocity (V2perpendicular) becomes zero.
Using the equation V2perpendicular = V1perpendicular + at, we can set V2perpendicular equal to zero since it stops momentarily before it starts descending. Solving for time:
0 = 25m/s + (-10m/s^2)(t_max)
Simplifying: 10m/s^2(t_max) = 25m/s
Solving for t_max: t_max = 25m/s / 10m/s^2 = 2.5s
So, it takes 2.5 seconds for the ball to reach its maximum height.
Note: Yes, the maximum height will be zero, as the ball stops momentarily before it starts descending.
To determine the maximum height above the ground, we can use the equation V2perpendicular^2 = V1perpendicular^2 + 2ad, where V2perpendicular is the final perpendicular velocity (which is zero at the maximum height), V1perpendicular is the initial perpendicular velocity (which we found to be 25m/s), a is the acceleration due to gravity (-10m/s^2), and d is the displacement or maximum height.
Using this equation, we can solve for d:
0^2 = (25m/s)^2 + 2(-10m/s^2)d
0 = 625m^2/s^2 - 20m/s^2d
20m/s^2d = 625m^2/s^2
d = 625m^2/s^2 / 20m/s^2
d ≈ 31.25m
So, the maximum height above the ground is approximately 31.25 meters.
To determine the position and velocity 3 seconds after being thrown, you can calculate the components of the velocity and displacement using the equations mentioned earlier.
The perpendicular component of the velocity (V2perpendicular) can be found using the equation V2perpendicular = V1perpendicular + at, where V1perpendicular is the initial perpendicular velocity (25m/s) and a is the acceleration due to gravity (-10m/s^2). Plugging in these values:
V2perpendicular = 25m/s + (-10m/s^2)(3s) = -5m/s (negative sign indicates downward direction)
For displacement, you can use the equation d = V1perpendicular(t) + 0.5at^2, where V1perpendicular is the initial perpendicular velocity (25m/s), t is the time (3s), and a is the acceleration due to gravity (-10m/s^2):
d = (25m/s)(3s) + 0.5(-10m/s^2)(3s)^2 = 60m
So, 3 seconds after being thrown, the ball's position is 60 meters above the ground and its velocity in the perpendicular direction is -5m/s (downward).