Range of Tennis Ball Thrown from Ground: 15 m/s

[DR33]

a tennis ball is thrown from the ground. 3 secs later, Vx = 15 m/s (horizontal). Determine the range.

 

[DR33]

Answer is 90.0m.. I just don't know how to get to it??

 

[NateTG]

If the horizontal velocity is 15 m/s and the ball spends 3 seconds in the air, then the total distance it travels is 45 meters.

Distance=rate*time.

If it takes 3 seconds to reach the peak of it's flight, then it goes 90m.

 

[DR33]

Thank so much..
I just made a fool of myself when I think about it again :p

L8er

 

[NateTG]

You asked about the angle of impact. It's easier to figure out the peak altitude by using:
$$\frac{1}{2}at^2$$

And realizing that the ball hits the top halfway through the fight.

If you want to find the angle of impact, the easiest option is that it's the same as the angle of launch. Alternatively, you can use the inverse tangent and the x and y velocities at impact.

 

[Julian Solos]

Originally posted by DR33
a tennis ball is thrown from the ground. 3 secs later, Vx = 15 m/s (horizontal). Determine the range.

You must have written the problem incorrectly. As it was written, it didn't have enough information.

"Vx = 15 m/s (horizontal)" was misleading. Vx is always horizontal, if we assume the x-axis is horizontal, and it's always 15 m/s if we are allowed to ignore air resistance.