Question

Solve a d earn 50 points.... Derive an expression for time of flight​

Answer 1

$\textbf{Answer is in Attachment !}$

Answer 2

hi mate,

Answer: The time of flight of an object, given the initial launch angle and initial velocity is found with: T=2visinθg T = 2 v i sin ⁡ . The angle of reach is the angle the object must be launched at in order to achieve a specific distance: θ=12sin−1(gdv2) θ = 1 2 sin − 1 ⁡ ( gd v 2 ) .

and......

An object thrown vertically up with some speed, its velocity will be decreasing and its velocity becomes zero when it attains the maximum height and then returns down with increasing velocity.If an object is thrown at angle to the vertical with some speed, its velocity will be decreasing and its velocity becomes zero when it attains the maximum height and then returns down with increasing velocity.The difference between the former and latter is that in the latter case it moves horizontally through some distance while duplicating the vertical fall of the former case.In terms of velocity, the object has two velocities one in the vertical direction and another in the horizontal direction.Thehorizontal velocitydoes not change in with time while thevertical velocity

Total time for which the object is in flight

It is denoted byT.

Total time of flight = Time of ascent + Time of descent

∴T=t+t= 2t Time of ascent = Time of descent =t]

At the highest pointH, the vertical component of velocity becomes zero. For vertical motion of the object (from 0 toH),

i hope it helps you.