A 0.19 kg, 0.26 m diameter red ball is thrown horizontally with an initial veloc

Question

A 0.19 kg, 0.26 m diameter red ball is thrown horizontally with an initial velocity of 18.9 m/s .
The ball was initially at the origin of the coordinate system at time zero.

Calculate the time, position vector, and the velocity vector for the ball when it is 39.1 m horizontally from its initial position.

time (__) s

position vector rvec = is correct(39.1) m i that + (__) m j that =(__) m angle (__) °

velocity vector vvec = is correct(18.9) m/s i that +(__)m/s j that = (__) m/s angle (__) °

need the rest of answer

Explanation / Answer

Hi, Since the ball was thrown horizontally, no vertical velocity is there but the ball will be acted upon by the gravitational pull downwards and hence it moves with constant velocity along the horizontal direction and under the acceleration due to gravity along -ve y-axis, assuming the ball thrown at origin (0,0) and towards +ve x-axis. Hence time taken to travel 39.1m horizonally       = horizontal distance / horizontal velocity = 39.1 / 18.9 = 2.069sec Distance the ball will fall vertically in this time is       s = ut + (1/2)at^2 = 0 - (0.5 * 9.8 * 2.069^2) = -20.976m Position vector = 39.1m i(hat) - 20.976m j(hat)
                         = sqrt(39.1^2 + 20.976^2)m with angle arctan(-20.976/39.1) deg with +ve x-axis                          = 44.372 m making an angle -28.21deg with +ve x-axis                          = 44.372 m making an angle -28.21deg with +ve x-axis   in this time the ball will gain a vertical velocity given by    v = u + at = 0 - (9.8 * 2.069) = -20.276 m/s Velocity vector = 18.9m/s i(hat) - 20.276m/s j(hat) (horizonal velocity is constant).                          = 27.719m/s with angle -47.012deg with +ve x-axis. Hope this helps you.

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.