Question

A small rubber ball is launched by a compressed-air cannon from ground level with an initial speed of 18.2 m/s directly upward. Choose upward as the positive direction in your analysis. (a) What is the maximum height above the ground that the ball reaches? 0 x What is the final velocity, momentarily, at the highest point? Knowing this, and the given information, can you use a constant acceleration kinematic relation to find the height? Also, be careful of signs. What is the sign of the initial velocity? What is the sign of the acceleration? m (b) How long does it take to reach the maximum height? 8.4 S (c) How long does it take to reach the ground after it reaches its highest point? S (d) What is its velocity just before it hits the ground? (Indicate the direction with the sign of your answer.) m/s

Answer 1

d) kinetic Now potential energy at top , PE = mgh at ground, kE= 1 mu? the velocity velocity of ball just before it hits ground. energy at ground where N is the So, migh = fmra gha ve 2 => v. pagh 12x 9.84 16.9 18.2 m/s V = -18.2m/s - v ball is falling down, so so negative sign is given As the ball is to v.

at top. Initial velocity u= 18.200/5 a) kinetic energy at ground - Potential energy Let top height ball beah. reeached by So, I mu² = mgh (182) 2 3. ५२ - => b= २१ 2x9.8 h b) So, = 16.9m maximum height reeached, h= 16.90 Let time taken to reach height, h=16. In be= t. h= ut + + gt² G = – 9.8m/s2 > 16.9 - 18.2t + (-4.9)t2 >> -4.9€ +18.27 -16.93 0 - 49+² + 182t - 169 = 0 = (-7t+13) (78-13) = 0 >> - 7t+13=0 -> t = 135 One 7t-13=0 Time taken to reach height = 16.94 = 7 This is the same time required to reach ground from highest point. time takes to Time taken to highest point . reach highest point from reach ground from ground- time, So 135. ta