Question

A projectile is launched with an initial speed of 21.8 m/s at an angle of 35º above the horizontal. Determine the time of flight of the projectile.

Answer 1

2.55 seconds

Step-by-step explanation:

Given that,

The initial speed of a projectile, u = 21.8 m/s

The angle of projection,
`\theta=35^(\circ)`

We need to find the time of flight of the projectile. The formula for the time of flight is given by :

`T=(2u\sin\theta)/(g)`

Where

g is the acceleration due to gravity

`T=(2* 21.8\sin(35))/(9.8)\\\\T=2.55\ s`

So, the time of flight of the projectile is 2.55 seconds.