Question

An object is launched at a velocity of 26 m/s in a direction making an angle of 33°

upward with the horizontal.
What is the horizontal range (maximum x above ground) of the object?
(use g = -10m/s2, and answer to the nearest 0.1m, but do not include units)

Answer 1

The horizontal range of an object launched at an angle can be found using the range formula, which incorporates the initial velocity, the angle of projection, and the acceleration due to gravity. Break down the initial velocity into horizontal and vertical components, then use these to find the time of flight and the horizontal range.

Step-by-step explanation:

To calculate the horizontal range of the object, we need to consider the initial velocity, the angle of projection, and the acceleration due to gravity. The given initial velocity is 26 m/s and the angle of projection is 33°. The acceleration due to gravity is -10 m/s² (negative sign indicates that gravity acts downwards).

Using the formula for the range of a projectile R = (v² ÷ g) × sin(2θ), where v is the initial velocity, g is the acceleration due to gravity, and θ is the angle of projection, we can calculate the horizontal range. The initial horizontal velocity (vᵀ) can be found using the equation vᵀ = v × cos(θ), and the initial vertical velocity (vₑ) can be found using vₑ = v × sin(θ). After finding the time of flight by calculating when the projectile will land back on the ground, we can use the horizontal velocity and flight time to find the maximum horizontal range.