a particle moves at a constant speed in a circular path, then the position vector (from the centre) and the instantaneous acceleration vector are:   Question 4 options:   both tangent to the circular path.   opposite in direction   perpendicular to each other.   in the same direction.   none of the above.

Question 22: At what angle a student must throw a ball in order to achieve the maximum ranget (a) 30° (b) 45° (c) 60° (d) 90° Question A body starts its motion from rest with an acceleration of 7 m/s. What is the final 23 velocity after 5 sec? 6 m/s (b) 15 m/s (c) 30 m/s (d) 35 m/s Question. The horizontal component of the acceleration (a.) for an ideal projectile motion is 24 equal to zero. (а) | True (b) False Question A projectile is launched with an initial velocity of 30 m/s at angle of 30° above the 25 horizontal. What is the magnitude of its velocity at t-2 secf. (a) 15.75 m/s (b) 10.2 m/s (c) 26.38 m/s (d) 2.2 m/s

How will the velocity change when the acceleration is perpendicular to it? None of the magnitude and direction will change. The magnitude of velocity will change and the direction will not. The direction of velocity will change and the magnitude will not. Both the magnitude and direction will change.

A projectile is fired with initial speed 150 m/s and an angle of elevation 45° from a position 10 m above ground level. Where does the projectile hit the ground and with what speed? Solution If we place the origin at ground level, then the initial position of the projectile is  (0, 10)  and so we need to adjust the parametric equations of the trajectory by adding 10 to the expression for y. With  v0 = 150 m/s,  ? = 45°, and g = 9.8 m/s2, we have x  =  150 cos    ? 4    t =          y  =  10 + 150 sin    ? 4    t −  1 2  (9.8)t2 =          . Impact occurs when y = 0, that is  4.9t2 − 75   2  t − 10 = 0.  Solving this quadratic equation (and using only the positive value of t), we get the following. (Round your answer to two decimal places.) t =  75   2  +    11,250 + 196 9.8  ≈   Then  x ≈ 75   2 (21.74) ≈    (rounded to the nearest whole number), so the projectile hits the ground about…