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
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 equationv*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
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