General physics question (position,velocity and acceleration)

[s4116792]

Hey I'm quite new to physics and am still getting my head around some equations and the correct ways to use them...

This question is just throwing me off because of the angle given and I don't know any formula's that include an angle.

Question is..

Steve and Mark are playing a game of backyard cricket. They set their field up so that the batsman stands 10.5metres from the point where the bowler releases the ball. Steve bowls the first ball of the match and the ball leaves his hand with a velocity of 60 km/h at an angle of 5 degrees below the horizontal. The ball is relased from a height of 2.1metres above the ground.

a) Where does the ball bounce
b) What is the speed of the ball just before it bounces?

I've drawn all the information out in a diagram just don't know what to do with it :S

Any help would be greatly appreciated..just need a point in the right direction with the equations :)

Thanks again

 

[Dick]

Your first step is to split the initial velocity into horizontal and vertical components. Horizontal is v0*cos(angle), vertical is v0*sin(angle). You should be learning that from that point on you can treat the horizontal and vertical motions independently.

 

[SWFanatic]

Using a diagram for this one is incredibly helpful (as you did). Using your trigonmic triangle to find the initial velocities of each plane(horizontal and vertical). The 60km/h should be the "hypotenuse of the vector diagram (triangle). Use the sin and cosine of 5 degrees to discover both components. From there, you can plug the new vertical initial velocity into the formula a = (vf-vi)/t where vf is the velocity is 0 because it has stopped moving upwards and is stopped for an instant, vi is the initial velocity, and a is the acceleration to Earth's gravity about -9.8 m/s^2(the ball is moving upwards). Find the time it takes to reach this point of 0 velocity(notice also that this is the point of maximum height. Next use on the formulas you have to find the distance traveled in that time.
Now do the same thing to discover how long it will take to travel to the ground (the distance you found to max height + the distance to the ground from which the ball was initially released. Remember the new vi is zero, since we are splitting this question up into two parts(its a little more understandable this way). Now you've found the total time it takes for the ball to transverse the vertical path (you must realize then that this is the same time that the ball will travel with horizontal speed. Since horizontal speed does not change (in this problem) use the simple formula v=change(d)/change(t).
Hope that helps!