a ball is thrown at 15m/s at an angle of 30 degrees to the horizontal , how far does it travel horizontally before making contact with the ground?
http://www.jiskha.com/display.cgi?id=1160269087
I will be happy to critique your thinking and work. Break the inital velocity into two components: vertical, and horizontal.
sorry..wrong quest.
If you are Jackie, stop changing names. It does not help us assist you. I noticed you and Jackie have the identical IP address. Thanks.
THAT'S LAME..MY SISTER AND I ARE USING THE SAME COMPUTER... BOTH HAVE HOMEWORK TO DO U KNO
Ok, I will assume then you don't read each others work here, unless I send you the links. Thanks. The previous problem she posted is a good example to the one you just posted. If she did it, her explaination to you would be helpful.
OK
i DID
S = VCOS THETA * 2VSIN (THETA)/G
= 13 * 2(7.5) / 9.81
= 195 / 9.81
= 19.9M
IS THAT KOOL?
Yes, your calculation is correct, but there is a slight mistake in your formula. The correct formula to calculate the horizontal distance traveled by the ball before hitting the ground is:
S = V₀ * cos(θ) * t
Where:
- S is the horizontal distance traveled
- V₀ is the initial velocity of the ball (15 m/s)
- θ is the angle of projection (30 degrees)
- t is the time it takes for the ball to hit the ground
To find the time, we can use the formula for vertical displacement:
y = V₀ * sin(θ) * t - (1/2) * g * t²
Where:
- y is the vertical displacement (height of the ball)
- g is the acceleration due to gravity (9.81 m/s²)
Since the ball hits the ground, the vertical displacement y is equal to zero. We can use this information to solve for t:
0 = (15 * sin(30)) * t - (1/2) * 9.81 * t²
Simplifying:
4.9 * t² = 7.5 * t
Rearranging and dividing both sides by t:
4.9t = 7.5
t = 7.5 / 4.9
t ≈ 1.53 seconds
Now we can substitute the value of t back into the formula for horizontal distance:
S = 15 * cos(30) * 1.53
S ≈ 19.9 meters
So, the ball travels approximately 19.9 meters horizontally before hitting the ground.