g=32.2ft/s²

r=unknown

Θ=0.5 * ((invsin)(r*g/v.²))

2Θ=((invsin)(r*g/v.²))

2Θ*v.²=((invsin)(r*g))

sin (2Θ*v.²)=r*g

Vy=15.96

Vx=12.036

therefore

sin (2Θ*v.²)/g=r

and

sin (2Θ*v.²)/r=g

——————–

first thing u do is look at the question

of course

u have 2 angles

1./ 53º

and

2./ 90º-53º=37º

you are given the resultant speed(hyp)

so now u have to find

the compotent speeds

there are 2

speed(Y)

and

speed(X)

so lets find both

53º above the horizon

therefore

20 * cos(53º)=speed(X) or 20 * sin(37º)=speed(X)

20 * sin(53º)=speed(Y) or 20 * cos(37º)=speed(Y)

so

20 * cos(53º)=speed(X) or 20 * sin(37º)=speed(X)

20 * 0.6018 =12.036

20 * sin(53º)=speed(Y) or 20 * sin(53º)=speed(Y)

20 * 0.798 = 15.96

now we have speed(Y) = 15.96

v= g * t

v = 32.2 * 0.4956

v = 32.2 * 0.4956

v =15.96 or speed(Y)

lets find distance to drop from the sky to ground

d(Y) = 0.5 * g * t²

d(Y) = 0.5 * 32.2 * (.4956)²

d(Y) = 7.97ft or distance(Y)

—————————

now u can calculate the Range

—————————-

d(X) = speed(X) * t

d(X) = 12.036 * .4956

d(X) = 5.965 ft

Carry on from here now

if u cant see this now

take up social science or politics LOL