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
It looks OK but I don't think you meant to say
"Note: maximum height will be 0, right?"
The vertical VELOCITY component at maximum height is zero.
With all those(4)significant figures, why are you using 10 for g instaed of 9.80 m/s^2 ? Unless they tell you do do that, I'd use a more accurate value. With four figures, it is about 9.804, but varies from place to place.
wooops I meant to say ask if the vertical velocity at maximum height is zero. And you answered that. Thanks! Yes, we were told to use 10 instead of 9.8.
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
Wait...it says location. How do I calculate the location? I only calculated it's vertical velocity.
I'm not sure if I did a right
but V2 x t
= 30.41 x 2
= 60.82m
The horizontal location at time t is
X = V1parallel * t
Th vertical location is V1perpendicular* t - (g/2)t^2
thanks!
d) is wrong, too. Can you help me with that?
To summarize:
a) To find the ball's location 2 seconds after being thrown, we need to calculate the parallel and perpendicular components of the velocity at that time. The parallel component remains constant at 30 m/s, and the perpendicular component can be found using the equation V2perpendicular = V1perpendicular + at, where a is the acceleration due to gravity (-10 m/s^2). Plug in the values to find that V2perpendicular is 5 m/s. Then, use the Pythagorean theorem to find the magnitude of the velocity V2 = sqrt(V2parallel^2 + V2perpendicular^2) = 30.41 m/s.
b) To determine the time it takes to reach maximum height, we can use the equation t = (V2perpendicular - V1perpendicular) / a, where a is the acceleration due to gravity (-10 m/s^2). Plug in the values to find that t = 2.5 seconds. The maximum height is reached when the vertical velocity component becomes zero.
c) The maximum height above the ground can be found using the equation V2perpendicular^2 = V1perpendicular^2 + 2ad, where a is the acceleration due to gravity (-10 m/s^2) and d is the maximum height. Rearrange the equation to solve for d: d = (V2perpendicular^2 - V1perpendicular^2) / (2a). Plug in the values to find that the maximum height is 31.25 meters.
d) To find the position and velocity of the ball 3 seconds after being thrown, we can use the equations V2perpendicular = V1perpendicular + at and d = V1perpendicular*t + (1/2)*a*t^2, where a is the acceleration due to gravity (-10 m/s^2). Plug in the values to find that V2perpendicular is -5 m/s and d is 60 meters. The negative sign indicates that the velocity is in the opposite direction of the initial velocity.