Question

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.

Answer 1

So,the above informations just depict a projectile motion,projected with velocity`u` 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`