The maximum height of an object, given the initial launch angle and initial velocity is found with:h=v2isin2θi2g h = v i 2 sin 2 θ i 2 g . The time of flight of an object, given the initial launch angle and initial velocity is found with: T=2visinθg T = 2 v i sin .
Formula:
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:
yan po ang sagot and mark brainlist and stay safe po
Answers & Comments
Answer:
The maximum height of an object, given the initial launch angle and initial velocity is found with:h=v2isin2θi2g h = v i 2 sin 2 θ i 2 g . The time of flight of an object, given the initial launch angle and initial velocity is found with: T=2visinθg T = 2 v i sin .
Formula:
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:
yan po ang sagot and mark brainlist and stay safe po