Ask your own question, for FREE!
Physics 20 Online
OpenStudy (abmon98):

A bomber flying to the right with a horizontal velocity of 600m/s at an altitude of hkm is to score a direct hit on a train moving to the left with a constant velocity of 180m/s and at a horizontal distance 15km from the plane as shown. 1) Calculate the angle a between the line of sight the target and the horizontal at the instant the bomb should be released? 2)The velocity when it hits the train? 3)The time it takes the missile to hit the train 4) The direction of the missile when it hits the train.

OpenStudy (abmon98):

For part 1) Should we use the trajectory equation and set y=0 since the missile started at the projection level Sx=vtcosθ work out t given the horizontal distance 15km then use vx=vcosθ and vy=vsinθ−gt. Taking the square root of v2x+v2y to find the velocity of the missile when it hits the train. my main doubt about this question is that the train is moving, if the missile hits the train then it wont be at a horizontal distance 15km

OpenStudy (anonymous):

Is the altitude of the plane given?

OpenStudy (abmon98):

nope

OpenStudy (abmon98):

it's given as h thats it

OpenStudy (irishboy123):

first find t for when the bomb and rain are at the same position in the horizontal direction, ie \(600t + 180t = 15000\) as they will collectively have covered the 15km as the bomb ie being dropped, its equation of motion in the vertical direction will be \(y = (0)t + \frac{1}{2} g t^2\). you know g and t. you can therefore calculate the height, h, that it **has to be at** for the train and the bomb to collide.

OpenStudy (abmon98):

@IrishBoy123 can we use relative velocity to be our initial speed of the bomb

OpenStudy (irishboy123):

sure you can.

OpenStudy (abmon98):

Sorry to bother you again if you dont mind checking those answers 1) y=xtan(theta)-g/2(x/x0)^2 0=15*10^3tan(theta)-5(15*10^3/600)^2 theta=11.768 2) relative velocity=600-(-180)=780m/s Horizontal velocity=780cos11.768 vertical displacement=-3125 sy=-1/2gt^2 t=25 seconds vertical velocity=-g(25) sqaure root of vx^2+vy^2 3) 25 seconds 4) tanx=vy/vx

OpenStudy (irishboy123):

not sure if this helps....

OpenStudy (abmon98):

I Appreciate your help thank you

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!