A ball thrown upwards vertically reaches a height 25 m in 1s. The ratio of total distance covered by the ball in time t=2s and t=4s is (use g = 10m/s2)
Given that, the height achieved in 1s,h=25m Let u be the initial velocity in upward direction. By equation of motion, h=ut−‌
1
2
gt2
25=u(1)−‌
1
2
(10)(1)2 25=u−5⇒u=30m∕s Final velocity of ball at maximum height become instantly zero. ∴ Time taken to reach at maximum height is calculated as v=u−gt⇒0=u−10t⇒t=‌
30
10
=3s Now, distance travelled in 2s= Displacement in 2s By equation of motion, d1=s1=ut−‌
1
2
gt2 =30(2)−‌
1
2
×10×(2)2 =60−5×4=60−20=40m Maximum height achieved by ball in (3s) H‌‌=ut−‌
1
2
gt2=30(3)−‌
1
2
×10(3)2 ‌‌=90−‌
10
2
×9 ‌‌=90−45=45m ‌ Displacement after ‌4s,‌ s2‌‌=ut−‌