Question

1. A projectile is fired straight upward at 170 m/s near the surface of Planet X, which has no atmosphere and where the acceleration due to gravity is g = 5 m/s2. (a) What is the acceleration of the projectile halfway to the top of its trajectory? [1] (b) Is the projectile in equilibrium at the top of its trajectory? Explain. [1.5] (c) How fast is it moving at the instant it reaches the top of its trajectory? (0.5] (d) How long does it take for the projectile to reach the top of its trajectory? (0.5) (e) Instead, suppose that it was fired at 170 m/s at an angle of 50° above a horizontal line. Then its vertical component of velocity is approximately 130 m/s, while its horizontal component of velocity is approximately 110 m/s. i. Why is the initial speed in the vertical direction larger than the horizontal speed? [1] ii. How long does it take for the projectile to reach the top of its trajectory in this case? (1) iii. What is the relative vertical position of the projectile (relative to where it was initially fired) 30 s after being fired? (1.5) iv. Accurately draw the horizontal and vertical velocity vector components at each of the points indicated in the trajectory of this projectile shown below. (A is the starting point and E is the ending point.) [3] B D AJ E

Answer 1

4 (a) a= 5m/52 (6) (6) At top of trajectory g. 5 m/s² pvia 170m/s ..The There is an acceleration in down and direction, The projectile is not at equilibrium attop- (C) at top V=0 at top of its trajectory Uf = vi-gt t = 345 = = 170-5t ottoms = 50° bg U Sinso'> Cossoo (Vy) = 170 sinso" (Ux); = 170 x cosso" Vyi> Uwi (6) at top of projectie, Vy so O= (y) : - gt 265 > t = 130 5 16som = ( Y = (y)it - £gt² = 130x30- 1 x 5 x 30² g (above the initial position)