sin(30) = 1/2
sin(37) = 0.6018150232
12.5/sin(30) = 25 meters
The takeoff for a 30 degree angle will depend with the offset angle.
1. You need to know the velocity of the projectile (V0) 2. The expressions for the range and height assume no air resistance (in vacuum) 3. The units must be consistent e.g. metres and g = 9.81 m/s2 Range in metres for 30 degree launch angle = sin 60 x V02 / 9.81 Range in metres for 45 degree launch angle = sin 90 x V02 / 9.81 Range in metres for 60 degree launch angle = sin 120 x V02 / 9.81 Max. height in metres for 30 degree launch angle = (V0 x sin 30)2 / 2g Max. height in metres for 45 degree launch angle = (V0 x sin 45)2 / 2g Max. height in metres for 60 degree launch angle = (V0 x sin 60)2 / 2g 2g is of course 9.81 x 2 = 19.62 m/s2 For interest, at 45 degree launch angle the max. height is 25% of the range.
sin(30) = 1/2
I assume your 90 degree angle is on the right and the 30 degree angle is opposite that. ( degree mode ) sin theta = opposite/hypotenuse sin 30 degrees = opp./44 = 22
5*sin(30) = 2.5 metres.
Use and rearrange the sine ratio: 30*sin(45) = 21.21320344 units
It is: 7.5*sin(30) = 3.75 meters
30 degree angle
sin(37) = 0.6018150232
Yes.
12.5/sin(30) = 25 meters
a 60 degree angle is twice the size of a 30 degree angle.
0.766
The takeoff for a 30 degree angle will depend with the offset angle.