Time = √(20 ÷ 4.9)
Horizontal distance = 50 * √(20 ÷ 4.9)
3. A golfer strikes the ball so the vertical component of its velocity is 20 m/s and the horizontal component is 30 m/s. How far will the ball traveled before it hits the ground what is the balls range? Remember your trick is to find total time in the air, for rise and for fall.
http://www.google.com/search?hl=en&sugex…
If you go to the website above, you will see that the path of the ball is an inverted parabola. A parabola is symmetrical.
Horizontal distance = horizontal velocity * total time
During the entire time the horizontal velocity is constant.
Horizontal distance = 30 * total time
As the ball rises, its vertical velocity decreases at rate of 9.8 m/s each second. At the highest position, its vertical velocity is 0 m/s. As the ball falls, its vertical velocity increases at rate of 9.8 m/s each second. The distance the ball moves upward is equal to the distance the ball moves downward. The magnitude of the initial vertical velocity is equal to the magnitude of the final vertical velocity. So the time for the upward trip is equal to the time for the downward trip.
So, total time = 2 * time up
Let’s determine the time for the upward trip.
vf = vi – 9.8 * t
0 = 20 – 9.8 * t
Time up = 20 ÷ 9.8
Total time = 2 * 20 ÷ 9.8
Horizontal distance = 30 * (2 * 20 ÷ 9.8)
The range is approximately 122 meters.
I hope this helps you understand projectile motion!