Question

A  gun with a muzzle velocity of 113.9 m/s is fired horizontally from a tower. Neglecting air resistance, how far downrange in meters will the bullet be 6.2 seconds later? Round your answer to 1 decimal place.

Answer 1

given initial horizontal velocity U(x) is 113.9 m/s and vertical velocity is U(y)=0 m/s

horizontal acceleration is A(x)=0 m/s and vertical acceleration is A(y)= g= 9.8 m/s

time=6.2 sec

horizontal case:

using the equation of motion   Range(x) = U(x)*t + 1/2*A(x)*t^2

Range(x)= 113.9*6.2 +1/2*0*6.2^2

Range(x)= 706.18 m

Vertical Case:

Range(y)= U(y)*t + 1/2*A(y)*t^2

Range(y)= 0*6.2 + 1/2*9.8*6.2^2

Range(y)= 188.356 m

This is the downrange

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