Juan throws a baseball with an initial speed of 40 m/s at an angle of 56 0 above the horizontal. Richard catches the ball at the same height that Juan threw it. How far did Juan throw the ball?
the basic range equation is
d = v^2 * sin(2Θ) / g
v = launch velocity
Θ = launch angle
g = gravitational acceleration
level surface (same height)
no air resistance
Scott you're a legend among men.
anyone who reads this, be sure you double angle theta in sine, it took me a second to figure that out, but no worries Scott thank you so much!!!!
To find the horizontal distance that Juan threw the ball, we can use the equations of motion for projectile motion. The horizontal and vertical motions are independent of each other.
We are given:
- Initial speed (magnitude of the velocity), v₀ = 40 m/s
- Launch angle, θ = 56° above the horizontal
To find the horizontal distance (range) traveled by the baseball, we can use the equation:
Range = (v₀² * sin(2θ)) / g
Where:
- v₀ is the initial speed (magnitude of velocity)
- θ is the launch angle
- g is the acceleration due to gravity, approximately 9.8 m/s²
Now let's substitute the given values into the equation to find the range:
Range = (40² * sin(2 * 56°)) / 9.8
To calculate this, we need to find the value of sin(2 * 56°). Double the angle first:
2θ = 2 * 56° = 112°
Now, substitute this value into the equation:
Range = (40² * sin(112°)) / 9.8
Using a scientific calculator or an online calculator, calculate sin(112°) to find the value and then solve the equation to determine the range.