Question

derive the formulas of
horizontal range
(oblique projectile motion )
​

Answer 1

Answer:

The range (R) of the projectile is the horizontal distance it travels during the motion. Using this equation vertically, we have that a = -g (the acceleration due to gravity) and the initial velocity in the vertical direction is usina (by resolving). Hence: y = utsina - ½ gt2 (1)

How to find the maximum height of a projectile?

if α = 90°, then the formula simplifies to: hmax = h + V₀² / (2 * g) and the time of flight is the longest. ...

if α = 45°, then the equation may be written as: ...

if α = 0°, then vertical velocity is equal to 0 (Vy = 0), and that's the case of horizontal projectile motion

Determine how high the projectile traveled above its initial height by using the following formula where V is the initial vertical velocity and T is the time it takes to reach its peak: Height = V * T +1/2 * -32.2 ft/s^2 *T^2 For example, if you had an initial vertical velocity of 32.14 ft/s and a time of one second

Answer 2

Answer:

The range (R) of the projectile is the horizontal distance it travels during the motion. Using this equation vertically, we have that a = -g (the acceleration due to gravity) and the initial velocity in the vertical direction is usina (by resolving). Hence: y = utsina - ½ gt2 (1)

Explanation:

Hopes it helps you