Question

•A ball is thrown upward from an initial height of   4.9 m with an initial speed of 9.8 m/s

•How high is the ball 2.0 s later?

•What is the velocity of the ball at its highest point

•When does it reach Its highest point

•How high does the ball go?

•When does the ball hit the ground?

•How fast is it going when (right before) it hits the ground?

Answer 1

^{we have,}    Initial height (h_i) = 4.9 m , Initial velocity (u_i) =9.8 m/s , g = 9.8 m/s²

• distance covered by ball after 2s (d) = u_i(2) - g(2)²/2 = 9.8*2 - 9.8*4/2 = 0 m

             ^therefore, height of ball after 2s later = h_i+d = 4.9m

• velocity of ball at highest point is u_f= 0 m/s
• time taken to reach highest point = (0-9.8m/s)/-9.8m/s² = 1s
• distance covered by ball in 1s i.e h_highest = 9.8*1 - g*1² /2= 9.8*1 - 9.8*0.5 = 4.9 m

          ^therefore, ball will go upto (4.9+4.9)m i.e. h_highest = 9.8 m

• 9.8 = 0*t + g/2 * t² = 4.9 t²

    ^so, t = 1.414 s

           ^therefore, ball will hit the ground after (1+1.414)s i.e. 2.414 s

• velocity before hitting the ground = v_f

         v_f² = u_f² + 2gh_highest = 0² + 2*9.8*9.8

         v_f = 13.86 m/s