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Launch angles closer to 45 ° 45\degree 45° give longer maximum horizontal distance (range) if initial speed is the same (see figure 5 above). These launches have a better balance of the initial velocity components that optimize the horizontal velocity and time in air (see figure 4).

The maximum height of an object, given the initial launch angle and initial velocity is found with: ｈ=v2isin2θi2g. The time of flight of an object, given the initial launch angle and initial velocity is found with: T=2visinθg .

For ideal projectile motion, which starts and ends at the same height, maximum range is achieved when the firing angle is 45°.

Answer and Explanation: 45 degrees is the optimal angle because it is the optimal combination of forward and upward components of velocity that give the projectile sufficient forward velocity and hang time to produce the greatest range.

So our equation for the launch angle of the projectile is 𝜃 equals the inverse sin of the square root of two 𝑔ℎ divided by 𝑉. All that's left to do is to substitute our known values of 𝑉, 𝑔, and ℎ into this equation to calculate 𝜃. The question tells us that the initial speed of the projectile is 28 meters per second.

The cannonball launched at a 45-degree angle had the greatest range. The cannonball launched at a 60-degree angle had the highest peak height before falling. The cannonball launched at the 30-degree angle reached the ground first.

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.

1:17 6:53 Projectile Motion - Calculating Launch Velocity - YouTube YouTube Start of suggested clip End of suggested clip Negative 30 is equal to U in the Y direction which is what we're trying to find. So this is a u inMoreNegative 30 is equal to U in the Y direction which is what we're trying to find. So this is a u in the Y. Direction.

Launch angles closer to 45 ° 45\degree 45° give longer maximum horizontal distance (range) if initial speed is the same (see figure 5 above). These launches have a better balance of the initial velocity components that optimize the horizontal velocity and time in air (see figure 4).

The maximum height of an object, given the initial launch angle and initial velocity is found with: ｈ=v2isin2θi2g. The time of flight of an object, given the initial launch angle and initial velocity is found with: T=2visinθg .

For ideal projectile motion, which starts and ends at the same height, maximum range is achieved when the firing angle is 45°.

Answer and Explanation: 45 degrees is the optimal angle because it is the optimal combination of forward and upward components of velocity that give the projectile sufficient forward velocity and hang time to produce the greatest range.

So our equation for the launch angle of the projectile is 𝜃 equals the inverse sin of the square root of two 𝑔ℎ divided by 𝑉. All that's left to do is to substitute our known values of 𝑉, 𝑔, and ℎ into this equation to calculate 𝜃. The question tells us that the initial speed of the projectile is 28 meters per second.

The cannonball launched at a 45-degree angle had the greatest range. The cannonball launched at a 60-degree angle had the highest peak height before falling. The cannonball launched at the 30-degree angle reached the ground first.

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.

1:17 6:53 Projectile Motion - Calculating Launch Velocity - YouTube YouTube Start of suggested clip End of suggested clip Negative 30 is equal to U in the Y direction which is what we're trying to find. So this is a u inMoreNegative 30 is equal to U in the Y direction which is what we're trying to find. So this is a u in the Y. Direction.