30 December 2023

16.The ratio of distance travelled by a freely falling body 2nd s and 3rd s is

Leave a Comment / Blog, Kinematics, Physics / By vibgyor4u.com

solution: For a freely falling body initial velocity (u)=0 for a body which travelled in nth second distance Sn=u+ a(n-1/2) for freely falling body acceleration(a)=g for 2nd second S2=g(2-1/2) =g(3/2) for 3rd second S3=g(3-1/2) =g(5/2) the ratio between is 3/2g:5/2g=3:5

15.An object projected vertically upwards with an initial velocity 40m/s from the ground(g=10m/s2).The time taken by the object to reach the maximum height

Leave a Comment / Blog, Kinematics, Physics / By vibgyor4u.com

solution: The object projected vertically upwards an initial velocity(u)=40m/s if the object to reach the maximum height, the final velocity will be zero. from kinematic equations ,we have the formula, (v)2-(u)2=2as here , v=0,u=40m/s, a=-g=10m/s2 0-40*40=2(-10)S -1600=-20S S=80m here, distance is nothing but maximum height .