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Vidushi Kaushal · Oct 15, 2012, 08:45 AM

Hey I need to know whether the answers I thought of for these questions are correct or not , so could you please answer them for me.

1) A ball is thrown horizontally at a place above the ground. Show in sketches ,

a) how the vertical velocity of the ball varies with time ,

b) how its horizontal velocity varies with time ,

c) the angle at which the ball strikes the ground of the vertical and horizontal velocities are then 20m/s and 5m/s respectively.

( No need to worry , you can just explain the shape of the graph.)

Curlyben · Oct 15, 2012, 09:13 AM

What do YOU think ?
While we're happy to HELP we won't do all the work for you.
Show us what you have done and where you are having problems..

Vidushi Kaushal · Oct 15, 2012, 09:32 AM

Umm , I think that the graph for horizontal velocity would be a straight line parallel to the time.

The graph for vertical velocity graph would be a straight line with constant but increasing gradient starting from 0 and inclined to the time axis.

The angle graph thing would be a curve with increasing gradient starting from 0 and ending at 75.6.

These were what I thought but I am very sure that they are wrong , so I needed to know what you had to say.

ebaines · Oct 15, 2012, 11:11 AM

Originally Posted by Vidushi Kaushal

umm , I think that the graph for horizontal velocity would be a straight line parallel to the time.

yes - constant horizontal velocity.

Originally Posted by Vidushi Kaushal

The graph for vertical velocity graph would be a straight line with constant but increasing gradient starting from 0 and inclined to the time axis.

I'm not sure what you mean by "constant but increasing gradient" - the vertical velocity graph would be linear, starting at 0 and heading down into the negative numbers with a constant slope.

Originally Posted by Vidushi Kaushal

the angle graph thing would be a curve with increasing gradient starting from 0 and ending at 75.6.

You are correct, but they're not asking for a graph in part c - they just want to know the angle of impact given Vh = 5m/s and Vv = -20m/s. You have the correct value for that.

Vidushi Kaushal · Oct 16, 2012, 05:32 AM

Thanks a lot! That was really helpful. Can you also tell me why the velocity time graph for the vertical velocity is heading down in to the negative numbers? Is it because we take the velocity downwards as negative and that upwards as positive?

ebaines · Oct 16, 2012, 05:43 AM

Originally Posted by Vidushi Kaushal

is it because we take the velocity downwards as negative and that upwards as positive?

Yes. You should get in the habit early in your study of physics to consistently use positive numbers for things that increase and negative numbers for things that decrease. For example If you throw a ball up in the air it starts with positive velocity (because it's moving upward, gaining altitude) but negative acceleration (because its positive velocity gets smaller as it rises), and after it reaches its peak and starts to fall it has negative velocity (because it's falling) and still negative acceleration (because it falls faster and faster).

Vidushi Kaushal · Oct 16, 2012, 08:08 AM

Thanks a lot. I have one question though , why won't the acceleration be positive when the ball is falling is the acceleration when the ball is rising is negative? Or does that only apply when doing problems based on the acceleration of free fall?

ebaines · Oct 16, 2012, 08:22 AM

Acceleration is defined as the change in velocity per unit time. Suppose you throw it upward with a positive velocity of 19.6 m/s. After one second it will have slowed to 9.8 m/s, so it's acceleration is:

$ a = \frac {\Delta V} {t} = \frac {V_2 - V_1} t = \frac {9.8 \frac m s - 19.6 \frac m s } {1 s} = -9.8 \frac m {s^2} $

Now consider its motion from t = 2 seconds to t = 3 seconds. At t=2 it reaches its max altitude and has v=0. From t=2 to t=3 it starts to fall, and reaches a velocity of -9.8m/s at t=3. Its acceleration is therefore:

$ a = \frac {\Delta V} {t} = \frac {V_2 - V_1} t = \frac {-9.8 \frac m s - 0 \frac m s } {1 s} = -9.8 \frac m {s^2} $

Again. It's negative. Under gravity the acceleration of an object in free fall (meaning only gravity is acting on it) is always -9.8 m/s^2, regardless of its velocity or direction of movement.

Vidushi Kaushal · Oct 16, 2012, 09:03 AM

Originally Posted by ebaines

Acceleration is defined as the change in velocity per unit time. Suppose you throw it upward with a positive velocity of 19.6 m/s. After one second it will have slowed to 9.8 m/s, so it's acceleration is:

$ a = \frac {\Delta V} {t} = \frac {V_2 - V_1} t = \frac {9.8 \frac m s - 19.6 \frac m s } {1 s} = -9.8 \frac m {s^2} $

Now consider its motion from t = 2 seconds to t = 3 seconds. At t=2 it reaches its max altitude and has v=0. From t=2 to t=3 it starts to fall, and reaches a velocity of -9.8m/s at t=3. Its acceleration is therefore:

$ a = \frac {\Delta V} {t} = \frac {V_2 - V_1} t = \frac {-9.8 \frac m s - 0 \frac m s } {1 s} = -9.8 \frac m {s^2} $

Again. it's negative. Under gravity the acceleration of an object in free fall (meaning only gravity is acting on it) is always -9.8 m/s^2, regardless of its velocity or direction of movement.

Thank you so much. Now only I understood the real mechanism.