How do you find a projectile's initial velocity?

Question

I have the following information
-Ranges from 0-15 feet
-Initial Trajectory Angles from 0 degrees to 90 degrees
-Acceleration is 32.15 ft/sec2 (and its due to gravity)

Please help me...I am so lost

Also...the highest range in feet is 15 feet and the initial trajectory angle is at 45 degrees

Answer 1

consider the initial velocity as a vector with two components Vx and Vy. Vx is constant as there is no acceleration in x (Not considering friction) Vx is variable, its acceleration is gravity. You can write an expression of the horizontal distance

distance = velocity x time
or
x=Vx * t

t = time the projectile is in the air. When the projectile reaches the earth it is the end of the process of course.

Now. Vox = Cos(a) * Vo

a = angle

Voy = Sin(a) * Vo

The vertical movement is accelerated, so acceleration is -g Vfy = 0 on the way up. The same on the way down (but in opposite direction and sign), so the total time the projectile is in the air equals to twice the time the pojectile needs to reach its maximum height from Voy=Sin(a) * Vo to Vfy=0 and can be expressed as:

Voy = g * t

or

t = Voy/g


Anyway So, if you have the projectile range of 15 feet at an angle of 45°, rearrange the equations above and use the following expression to obtain Vo

x=(Vo2/g)Sen(2*a)

x=15 ft
g = 32.15 ft/sec2
a = 45°

Hope it helps

Answer 2

D, seeing that the formulation is like this V=Vo+gt shall we see the moment facet raises as instances passes, thus for those who analise the dimensionals for this moment facet, they end up an growing pace g=m/sec^two t= sec m/sec^2times sec= m/sec the dimensionals for pace