How do I solve for initial velocity with the given data? ( Kinematics questions)

45 Degrees Angle
Height of Y = 1.025 M
Displacement of X = 1.397 M
Displacement of Y = 0 M
Acceleration of X = 0 m/s^2
Acceleration of Y = -9.8 m/s^2

I got 2.8 for my initial velocity, but I don't know if it is right.

1 Answer
Feb 28, 2018

So,the above informations just depict a projectile motion,projected with velocityu at an angle 45 w.r.t horizontal,so,vertically it went upto a maximum height(1.025m) and then came down,so displacement (Y) is zero,and its range of motion is 1.397m

So,we can write,

1.379=ucos45T (using, s=vT for horizontal motion, where, T is the total time of flight)

Now,total time of flight for a projectile motion is twice the tie required to reach the highest point,

So, using v=ugt we can write, 0=usin45gt(as,at highest point vertical component of velocity =0)

So, T=2t=2usin45g

So, ucos45=1.397g2usin45

And,using v2=u22gH we get, 02=(usin45)22g1.025

So, u2sin245=2.05g or, usin45=4.47

So,ucos45=1.53 (putting the value of usin45 in the previous equation)

So, u2sin245+u2cos245=u2=(2.05g)+(1.53)2

or, u=4.738ms1