Table of Contents
Why does 45 degrees give the maximum range?
The sine function reaches its largest output value, 1, with an input angle of 90 degrees, so we can see that for the longest-range punts 2θ = 90 degrees and, therefore, θ = 45 degrees. A projectile, in other words, travels the farthest when it is launched at an angle of 45 degrees.
What is the maximum height that the ball reaches?
Question: A ball thrown vertically upward reaches a maximum height of 30 meters above the surface of Earth. At its maximum height, the speed of the ball is: Answer: 0 m/s. The instantaneous speed of any projectile at its maximum height is zero….Objects Launched Upward.
| Variable | Value |
|---|---|
| a | -9.8 m/s2 |
| t | 0.40 s |
Why is a 45 degree best angle for launch?
Originally Answered: Why is 45 degrees the optimal angle for projectiles? It maximizes the projectile’s range (the horizontal distance the projectile travels). Aiming too high causes the projectile to fly higher into the air at a reduced range.
What is the initial horizontal velocity of a pool ball?
To illustrate the usefulness of the above equations in making predictions about the motion of a projectile, consider the solution to the following problem. A pool ball leaves a 0.60-meter high table with an initial horizontal velocity of 2.4 m/s.
What is the formula for launching a projectile from a height?
Fortunately in the case of launching a projectile from some initial height h, we need to simply add that value into the final formula: hmax = h + V² * sin (α)² / (2 * g)
How do you determine the initial velocity of a soccer ball?
Determine the initial horizontal velocity of the soccer ball. A projectile is launched at an angle to the horizontal and rises upwards to a peak while moving horizontally. Upon reaching the peak, the projectile falls with a motion that is symmetrical to its path upwards to the peak.
How do you find the range of a projectile motion?
if α = 45°, then the equation may be written as: hmax = h + V₀² / (4 * g) and in that case, the range is maximal if launching from the ground (h = 0). if α = 0°, then vertical velocity is equal to 0 (Vy = 0), and that’s the case of horizontal projectile motion.