Q1 )A projectile thrown with a velocity 49m/s in the upward direction , making an angle of 30 degree with the horizontal . Find the velocity after 2 seconds.(g=9.8m/s)















Q2)Two guns are used to fire 2 bullets of masses 50 g & 75 g , with horizontal speeds of 100 m/s and 125 m/s respectively , from a height of 20 m above the ground . Find the ratio of the time intervals taken by the bullets to reach the ground.















Q3)A body travels 200 cm in the fiest 2 seconds and 220cm in the next 4 seconds , what will be the velocity at the end of the 7th second ?















Q4) One body is dropped while the second body at the same instant is thrown downward with an initial velocity of 1m/s . When will the distance between them be 18 m?

3)ans) let u n a be the intial velocity n acceleration of the body for 2sec distance covered = 200cm ,

usin s=ut+1/2at² = v have 200= u(2) +(1/2)a(2)² ---- (1)

n after 4 sec of the journey i.e after a time t = 6sec the distance covered = 200+220 = 420cm

hence 420 = u(6) +(1/2) a(6)² ----(2)

solvin (1) n (2) v have  u = 115 cm/sec n a= -15 cm/sec² 

now velocity after 7 sec = u+at = 115+(-15)7 = 10cm/sec 

rate if useful  

4)ans) since both bodies r fallin frm the same height under influence of gravity , so relative acceleration of one w . r. to each other is ZERO ..

so h =  . 

=> = t = 18 / 1-0 = 18 sec

rate if useful  

2) time taken =rt(2h/g)

since h is same ............den ratio 1:1........

(1). Its horizontal velocity will be 49root3 / 2 m/s ( resolving the components of velocity )

       After 2s, horizontal velocity will b same...as it is does not change while projection.

       Its vertical velocity will be 49/2 = 24.5 m/s

       After 2s, vertical velocity will be.........  v = u - gt

                                                                            = 24.5 - 9.8 (2)

                                                                            = 24.5 - 19.6  =  4.9 m/s

       Now, taking resultant of both the velocities......

              Velocity at 2s = 

Correct me,if I am wrong.

HERE REPLY OF 3rd and 4th

(3)

now distance travelled in first 2 sec = 200cm

let u be initial vel

so eqn =

and in first 4 sec distance = 4.2m

so eqn =

solving eqn  a = 0.05 and u = 0.95

so distance at 7th sec= 7u + 24.5a = 7.8m

(4) now for body dropped

eqn = s =

and for other part

s =  u = 1

so difference = 18m after 18sec

plz rate yaaaarrrrrrrr!!!!!!!!1

best of luck

this Q can be easily handled with eq of trajectory with this eq you can find every thing in a single point of the path

of projectile and trajectory means path followed by the particle in the projectile

here is this

y=Xtan@-(gx^2)/(2u^2cos^@) and where @ is the angle of the projectile acc to the Q

various forms of the trajectory is

y=xtan@(1-x/R)

y=xtan@-gx^2(1+tan^2@)/(2U^2)

and the practise makes a man perfect i hope this info is useful to you