a bullet is shot with initial speed of 60m/s at angle of 30,it would travel approximately?

Answer 1

OK, if the angle is 30 degrees up from horizontal, then the "vertical" component of the speed is sin(30)*60 m/s.
Sin(30) = .5, so the bullet is going "up" at 30 m/s.
The acceleration of gravity is -9.8 m/s/s, so the bullet will be going UP for 30/9.8 seconds, and will then go DOWN for 30/9.8 seconds, for a total time in the air of 60/9.8 seconds, or 6.12 seconds.

During that 6.12 seconds, you might want to know how FAR the bullet will travel. Well, for this you need to know the "horizontal" component of the bullet's speed, which is cos(30 degrees)*60 m/s, = .866*60 m/s
Multiply this by the amount of time in the air, 6.12 seconds, and you get the distance = .866*60*6.12 meters, or 318 meters.

The assumptions in this are that air friction is negligible, and that the height of the shooter is small.

Answer 2

Using the formula of horizontal range u^2 sin2a / g where a is the angle of projection and u is the initial speed. We get 60x60xsin60 / 10

Final answer is 311.76m

The answer will be more accurate if we take g=9.8m/s^2 but in the above question, it has been taken as 10m/s^2

Answer 3

sin(30)60m/s= 30m/s
cos(30)60m/s= 52m/s

V=at+Vi
-30m/s=(-9.8m/s^2)t+30m/s
-60m/s=(-9.8m/s^2)t
t= 6.1s

x=vt
x= (52m/s)(6.1s)
x=317.2m