# Answer to Question #4514 in Mechanics | Relativity for Mitch

Question #4514

You shoot a projectile at an unknown angle but at a velocity of 15m/s. You know that this must hit a target 20 feet away (height of the target negligible). What is the time of the flight and how do you get it.

Expert's answer

Let Alpha be the angle between the direction of the shoot and the horizont, V=15m/s is the initial velocity. Let T be the time while the projectile flies.

Then

S = V*cos(Alpha)*T - equation along the horizontal axis (S = 20 feet).

V*sin(Alpha) = g*T/2.

From the first equation (cos(Alpha))^2 = (S/(V*T))^2, from the second (sin(Alpha))^2 = (g*T/(2V))^2.

As (sin(Alpha))^2+(cos(Alpha))^2 = 1,

then:

(S/(V*T))^2+(g*T/(2V))^2=1

4S^2+g^2*T^4 = 4V^2*T^2

g^2*T^4 - 4V^2*T^2 + 4S^2=0 - biquadratic equation

D = 16*V^4-16*S^2*g^2* = 16*(V^4 - S^2 * g^2)

So, T = sqrt((4*V^2-4*sqrt(V^4 - S^2 * g^2)/(2*g^2)).

T = sqrt(2*(V^2-sqrt(V^4 - S^2 * g^2)/g^2).

T = sqrt(2*(15^2-sqrt(15^4-6.01^2*9.8^2))/9.8^2) = 0.4 seconds.

Then

S = V*cos(Alpha)*T - equation along the horizontal axis (S = 20 feet).

V*sin(Alpha) = g*T/2.

From the first equation (cos(Alpha))^2 = (S/(V*T))^2, from the second (sin(Alpha))^2 = (g*T/(2V))^2.

As (sin(Alpha))^2+(cos(Alpha))^2 = 1,

then:

(S/(V*T))^2+(g*T/(2V))^2=1

4S^2+g^2*T^4 = 4V^2*T^2

g^2*T^4 - 4V^2*T^2 + 4S^2=0 - biquadratic equation

D = 16*V^4-16*S^2*g^2* = 16*(V^4 - S^2 * g^2)

So, T = sqrt((4*V^2-4*sqrt(V^4 - S^2 * g^2)/(2*g^2)).

T = sqrt(2*(V^2-sqrt(V^4 - S^2 * g^2)/g^2).

T = sqrt(2*(15^2-sqrt(15^4-6.01^2*9.8^2))/9.8^2) = 0.4 seconds.

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