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If a ball is thrown at a velocity of 43ms at an angle of 32degrees above the ground what is the ball's initial velocity component of velocity Vv0 ---- How long will the ball be in the air?
Wiki User
ā 16y ago
Vv0 =sin(32)*43 m/s
=22.78 m/s
h = Vv0*(t) + 0.5gt^2
if we assume g=10 m/s^2
when the ball lands, h=0
0=22.78*t + 5(t)(t)
t(22.78+5t)=0
t={0s,4.56s}
The ball will be in the air for about 4.5 seconds
Wiki User
ā 16y ago
The initial vertical component of velocity, Vv0, can be calculated as: Vv0 = 43 * sin(32Ā°) ā 22.27 m/s. The time the ball will be in the air can be determined using the kinematic equation: time = 2 * Vv0 / g, where g is the acceleration due to gravity (9.81 m/sĀ²). Substituting the values, time ā 4.56 seconds.
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