A baseball is thrown with a velocity of 31m/s at an angle of40 o from a height o

Question

A baseball is thrown with a velocity of 31m/s at an angle of40o from a height of 1.8m. Ignoring the effect of airresistance, what will be: A) the horizontal velocity of the ball? B) the vertical velocity when the ball returns to the heightit was thrown? C) the flight time until the ball returns to the height it wasthrown? D) the time to apex (peak height) of the ball's flight? E) the height of the ball at the apex? F) the distance down field that the player needs to catch theball at the height it was thrown? A baseball is thrown with a velocity of 31m/s at an angle of40o from a height of 1.8m. Ignoring the effect of airresistance, what will be: A) the horizontal velocity of the ball? B) the vertical velocity when the ball returns to the heightit was thrown? C) the flight time until the ball returns to the height it wasthrown? D) the time to apex (peak height) of the ball's flight? E) the height of the ball at the apex? F) the distance down field that the player needs to catch theball at the height it was thrown?

Explanation / Answer

given the velocity V = 31 m/s the angle is = 40o a) the horizontal velocity is = V cos40o =23.7473 m/s b) the verctical velocity is = V sin 40o=19.9264 m/s c ) the flight time is T = (2V sin) / g = 4.066 s d) the time to apex = T /2 = 2.0333 s E ) the maximum height is H = (V2sin2 ) / 2g c ) the flight time is T = (2V sin) / g = 4.066 s d) the time to apex = T /2 = 2.0333 s E ) the maximum height is H = (V2sin2 ) / 2g

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