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# 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.
1
2011-10-11T12:32:35-0400
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.

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