1. ## Projectiles

If a bullet is fired from a gun at a point o with a speed of 150m/s at an angle Tan-1(3/4) to the horizontal.At the instant of firing,a moving target is 25.1 metres vertically above o and is moving at an angle of 45 degrees to the horizontal at a constant speed of 90root2 m/s.show that the bullet strikes the target after 1 second.where does the strike occur???assume there on the same plane.

i got 85.1 ,but i didn't use all the info given so im not confident with the answer.Can anybody tell me what they got and how?

2. The horizontal initial velocity is

$v_x(0) = v(0)\cos(\theta) = 150(.6) = \text{ 90 m/s }$

and the vertical initial velocity is

$v_y(0) = v(0)\sin(\theta) = 150(.8) = \text{ 120 m/s }$.

The horizontal velocity remains the same throughout the bullets flight since there is no acceleration in the x-direction after the bullet is fire. But the velocity in the y-direction is changing because of the acceleration due to gravity. Initially this is downward (in the opposite direction of the vertical velocity) and the velocity as a function of time is governed by the equation: $v_y(t) = v_y(0) a t$ where $a = -9.81m/s^2$.

Also the x-position of the bullet is $x(t) = v_x(t)t$ while the y-position is $y(t) = v_y(0)t + 1/2at^2$.

So after 1 second the bullet has traveled $x(1) = v_x(1)1 = 90 m/s (1 s) = 90m$ in the x-direction and $y(1) = 90m/s(1 s) - 9.81/2(1)^2 = 115.095m$ in the y-direction. So you might say that the bullet is at the position (90,115.1)

Now the target is moving $90 m/s$ in both the x and y directions. The speed is constant. so after 1 second the target has moved 90m vertical and 90m horizontal putting its position after 1 second at exactly (90,115.1), which is where the target is.