A shell fired from a gun has a horizontal range of 1200 m and a flight time of 10 s. What is the magnitude and direction of the velocity of projection?

2 Answers
May 14, 2018

#|v| = 129.64 m/s#
#theta = 22.23°#

Explanation:

#x = v_{0x} * t = v_{0x} * 10 = 1200#
#=> v_{0x} = 120 m/s#
#v_y = v_{0y} - g*t#
#"At the highest point "v_y = 0" and "t = 10/2 = 5 s#
#=> 0 = v_{0y} - 9.81*5#
#=> v_{0y} = 49.05 m/s#

#=> |v| = sqrt(v_{0x}^2 + v_{0y}^2) = 129.64 m/s#
#theta = arctan(v_{0y}/v_{0x}) = 22.23°#
#"So the direction is 22.23° above the horizontal ground."#

May 14, 2018

The velocity of projection is #=129.6ms^-1# and the angle is #=22.2^@#

Explanation:

The range is

#r=v^2sin(2theta)/g#

The time of flight is

#t=2vsintheta/g#

Therefore,

#v^2sin(2theta)/g=1200#................#(1)#

and

#2vsintheta/g=10#........................#(2)#

So,

Equation #(1)# divided by #(2)# is

#vcostheta=120#

The acceleration due to gravity is #g=9.8ms^-2#

From #(2)#

#vsintheta=5g=49#

#=>#, #(vsintheta)/(vcostheta)=49/120#

#tantheta=0.4083#

#theta=arctan(0.4083)=22.2^@#

#v=120/costheta=120/cos(22.2^@)=129.6ms^-1#