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# Magnitude of the initial velocity?

label Physics
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A cannon sends a projectile towards a target

a distance 1540 m away. The initial velocity

makes an angle 37◦ with the horizontal. The

target is hit.

The acceleration of gravity is 9.8m/s/s

What is the magnitude of the initial velocity? Answer in m/s

Nov 17th, 2017

Let v be the unknown speed. We will deduce distance in terms of v, make it equal to 1540 and solve for v.

The position of a projectile at time t in a vector form may be written as
r=<v*cos(37*pi/180)*t , v*sin(37*pi/180)*t - (9.8*t^2)/2>

Making the second component equal to zero, we find t>0 at which projectile reaches the ground.
v*sin(37*pi/180)*t - 4.9*t^2 = 0,  so
t=v*sin(37*pi/180)/4.9.

Substituting this into firs component of position and making it equal to 1540 we get equation
v*cos(37*pi/180)*(v*sin(37*pi/180)/4.9)=1540 or

(cos(37*pi/180)*sin(37*pi/180)/4.9)*v^2 = 1540

solving for v we get

v=1540*4.9/(cos(37*pi/180)*sin(37*pi/180))  ~  125.30 m/s

Oct 24th, 2014

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Nov 17th, 2017
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Nov 17th, 2017
Nov 18th, 2017
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