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= u cos 45 *T (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=u - g t we can write, 0=u sin 45 -g t(as,at highest point vertical component of velocity =0)

So, T =2t = (2u sin 45)/g

So, u cos 45 = (1.397g)/(2u sin 45)

And,using v^2=u^2 -2g H we get, 0^2 =(u sin 45)^2 - 2g1.025

So, u^2 sin ^2 45 = 2.05g or, u sin 45 =4.47

So, u cos 45 =1.53 (putting the value of u sin 45 in the previous equation)

So, u^2 sin^2 45 + u^2 cos ^2 45 = u^2 =(2.05g) +(1.53)^2

or, u=4.738 ms^1