Projectile motion of a ball launcher?
A ball is thrown out of a ball launcher projectile. If the ball lands in a pile of books (same height as ball launcher) 8 m away and the ball was in the air for 2.5 seconds, which is the angle above the horizon as the ball is launched? Really almost have no idea how to start from this problem. : / (B) What was the total initial velocity, and (c) how high above books / launcher ball traveled the ball?
The ball goes from 8 m in 2.5 s, so the horizontal velocity component VX0 = 8/2.5 = 3.2 m / s The ball reaches maximum height in half the total travel time and the maximum height occurs when the vertical velocity is 0. Vertical velocity is Vy0 – t * g, when this is 0, Vy0 g = t *, so Vy0 = 9.8 * 1.25 = 12.25 m / s, the launch angle is arctan (Vy0/Vx0) = arctan 12.25/3.2 = 75 ° The initial velocity is √ [Vy0 VX0 ² ² +] = 12.7 m / s vertical height is h = t * Vy0 – 0.5 g * * ² t, where t = 1.25 h = 7.7 m
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