Solving Steve and Mark's Backyard Cricket Problem

[s4116792]

Hey this is my second post on this subject sorry to bother but i still just can't get my head around it...i have used so many different equations and the answers just don't seem to be correct.

The 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 got 4.8 metres distance which just doesn't seem likely to me and 19.3 m/s

The next part of the question just changes the angle to 3 degrees above horizontal and asks the same questions, this bit is really throwing me off because all of my textbooks only have equations 'from origin' and the ball leaves 2.1 meters above the ground and goes up then down to 0 meters.

If someone could please explain this to me very simply it would be very greatly appreciated...this is my first year of physics :)

 

[Integral]

Please do not start multiple threads on the same question. Your first post was moved to the homework help forum and is https://www.physicsforums.com/showthread.php?t=221257" Please continue your questions in that thread.

Thanks.

 

[charng]

I understand your frustration and difficulty in solving this problem. It can be challenging to apply equations to real-world scenarios, especially when the situation is constantly changing.

Firstly, it is important to double-check your equations and calculations to ensure they are correct. If you are still getting an answer that seems unlikely, it may be worth seeking assistance from a classmate or your teacher.

In terms of the question, it is important to remember that the ball will follow a parabolic path due to gravity. This means that it will rise to its highest point and then fall back down to the ground. The equations you have may be for projectile motion from a fixed point, but you can still apply them to this scenario.

In terms of the first part of the question, to find where the ball bounces, you can use the equation for the vertical displacement of a projectile:

y = y0 + v0y*t + (1/2)*a*t^2

Where:
y = vertical displacement
y0 = initial vertical position (2.1m)
v0y = initial vertical velocity (in this case, 60km/h converted to m/s)
a = acceleration due to gravity (-9.8m/s^2)
t = time

You can solve for t by setting y equal to 0 (since the ball will hit the ground at this point). This will give you the time it takes for the ball to hit the ground. Then, you can use this time in the horizontal displacement equation:

x = x0 + v0x*t

Where:
x = horizontal displacement
x0 = initial horizontal position (0m)
v0x = initial horizontal velocity (in this case, 60km/h converted to m/s)

This will give you the distance the ball travels before it bounces. Remember to convert the units to meters for consistency.

For the second part of the question, where the angle is changed to 3 degrees above horizontal, you can use the same equations but with a different initial vertical velocity (since the angle has changed). You can also use the Pythagorean theorem to find the initial horizontal velocity.

I hope this helps to simplify the problem for you. Remember to always double-check your work and seek assistance if needed. Good luck with your studies!