How Does Angle Affect the Acceleration of a Thrown Ball?

[spyroarcher]

Homework Statement

A person throws a 175 gram ball by exerting a force of 10N, 45 degrees.
a) Determine the resulting acceleration of the ball.
b) Repeat for the same amount of force applied by the girl in a downward direction.

Homework Equations

F=ma

The Attempt at a Solution

For a, I simply did a=10/.175, but I suspect that it is wrong since the angle at which it was thrown can change that answer and gravity plays some role...
For b, I don't quite understand what is happening, like does the person throw at an angle of -45 degrees now? Do I also assume the person throwing is still throwing at 10N, since I think it is influenced by gravity...
Thanks in advance

 

[fss]

Draw a free body diagram and use geometry to separate the acceleration vectors into components.

 

[94JZA80]

for part a, you'll want to break up the acceleration induced by the 10N force into horizontal and vertical components. the acceleration you calculated (a = F/m = 10/0.175 = 57.2m/s²) is the amount of acceleration experienced by a 175-gram object being acted upon by a 10N force IN THAT SPECIFIC DIRECTION 45° from the horizontal. you can find the vertical and horizontal components of that acceleration by treating 57.2m/s² as the hypotenuse of a 45-45-90 triangle, and solving for the lengths (components of acceleration) of the other sides of the triangle using the definition of sine or cosine...of course you really only have to solve for one of them b/c the short sides of a 45-45-90 triangle are equal in length. at this point, you can combine the vertical (specifically, upward) component of acceleration induced by the 10N force with the other vertical (specifically, downward) component of acceleration you need be concerned with - the acceleration induced by gravity. now you can use this new vertical component of acceleration (which is the sum of the accelerations induced by the vertical component of the 10N force and gravity itself) along with the horizontal component of acceleration to create a new right triangle. again, using the definition of sine or cosine, you can calculate the length of the hypotenuse (the net acceleration induced by both the 10N force AND gravity), which is your answer for part a. alternatively, you can start by breaking up your 10N force in the 45° direction into horizontal and vertical force components (instead of first calculating the acceleration and breaking that up into components). you can then combine the vertical (upward) component of the 10N force with the vertical (downward) force of gravity on the ball. this force would be 0.175kg x 9.8m/s² = 1.715N in the downward direction. again, by creating a new right triangle with the net vertical force component and the horizontal force component, you can use the definition of sine or cosine to find the length of the hypotenuse (the net force created by combining the 10N force and the force of gravity), and then calculate acceleration by using a = F/m.

as for part b, I'm also confused as to what the question is really asking. is the person now throwing the ball 45° downward instead of upward this time? or are you referring to the downward force of the person's feet on the ground as they throw/force the ball upward?

 

[NeedsHelp1212]

as others have said draw a free body diagram. Label all the forces involved, and draw components for the x and y. I have never covered this yet in my physics class where the ball is thrown, this is much more complicated since i think there's a net force in both the -y direction and +x direction. Right now we simply have objects going down ramps, or situations with pulleys, not projectile motion.

 

[94JZA80]

as others have said draw a free body diagram. Label all the forces involved, and draw components for the x and y. I have never covered this yet in my physics class where the ball is thrown, this is much more complicated since i think there's a net force in both the -y direction and +x direction. Right now we simply have objects going down ramps, or situations with pulleys, not projectile motion.

actually the net force in part a (where the ball is thrown upward at a 45° angle with the horizontal) in the y-direction is positive - at least for the short moment that the 10N force is being applied to the ball. after that, only the force of gravity is left acting on the ball, so from that point on, yes - the net force in the y-direction is negative. it is the reason the ball immediately begins to decelerate in its upward direction the instant it leaves the person's hand, eventually stops its vertical motion altogether for an instant, and then begins accelerating again toward the ground. if the net force in the y-direction were negative from the start, the ball would have never initially traveled upward, and it would have left the hand of the person in a downward direction.

 

[venkatg]

Assuming that the experiment is done on earth, we include gravity.

The Earth gravity is pointing down.

The 10N force that is angled 45 deg can be resolved into vertical and horizontal components

The net vertical component is

V = (175*9.81 - 10*sine(45)). This will be pointing down.
acceleration is A1 = V/M

The horizontal component of the force is
H = 10 * cos(45).
acceleration A2 = H/M

The resultant is A = sqrt(A1^2 + A2^2)