60 649
Assignments Done
Successfully Done
In April 2018

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.
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,


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.

Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!


No comments. Be first!

Leave a comment

Ask Your question

Privacy policy Terms and Conditions