A projectile is launched with a velocity of 21.9 m/s at an angle of 62.9 deg abo
ID: 1953460 • Letter: A
Question
A projectile is launched with a velocity of 21.9m/s at an angle of 62.9 deg above the ground.The projectile lands at the same elevation that it was launched from. While the projectile is going up, when it reaches 1/4 of its maximum height, what is the angle of its velocity vector relative to the ground? (deg, three significant figures, Use 9.8m/s2 as the acceleration due to gravity) A projectile is launched with a velocity of 21.9m/s at an angle of 62.9 deg above the ground.The projectile lands at the same elevation that it was launched from. While the projectile is going up, when it reaches 1/4 of its maximum height, what is the angle of its velocity vector relative to the ground? (deg, three significant figures, Use 9.8m/s2 as the acceleration due to gravity)Explanation / Answer
Horizontal vellocity remains constant
v = 19.61 m/sec
Thus tan = 16.87/10 = 59 . 23 degree
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