Physics help needed kinetics?

Question

1) you are in a hot air balloon relative to the ground has a velocity of 6.0 m/s due east. you see a bird moving directly away from the balloon due north. The speed of the hawk relative to you is 2.0 m/s. What are the magnitude and direction of the hawks velocity relative to the ground?Express the angle relative to due east.

2)A golfer tees off and drives the ball 3.5 times as far as he would n earth, given the same initial velocities on both planets. The ball is launched at 45 m/s at an angle of 29 degrees above the horizotal. When the ball lands it is the same level as the tee. On the distance planet what are the aximum height and range of the ball?

i appreciate any help i can get, please show your work even if you can only help me with 1 of the 2 problems. thanks so much!

Answer 1

There is a little bit of ambiguity in the question that leaves it open to interpretation, but let's see if I can lend you a hand.
In reading the first question, I am in a balloon that is moving due east at 6m/s. A bird (in this case a hawk) is flying in an unknown direction, but his position from me is moving north at a rate of 2m/s.
The bird is flying roughly east by northeast. The cosine of his angle from due east is represented by the vector of the balloon, 6m/s. The sine of his angle from due east is represented by the vector he is moving north, 2m/s. Since the tangent of that angle is equal to the sine/cosine, then 2/6 is the tangent of that angle and therefore the arc tangent of 2/6 is that angle relative to east.

And now for the second problem. The key is in the info that the range is 3.5 times as far as it would be on earth. That tells us that the ball spent 3.5 times as much time in the air. It went upward 3.5 times longer and then it went down ward 3.5 times longer. (that is NOT 7 times longer, but 7 times 1/2 longer)
So the accelleration due to Gravity is 3.5 times weaker or g/3.5 where g=9.8m/s^2.
At this point the problem is similar but slightly more complex than the first one. There are four elementary equations that can be used to solve most kinematic problems.
A: d=vo*t+(a*t^2)/2 where vo is initial velocity, t is time and a is accelleration or in this case gravity and is a negative value.
B: vf^2=(vo^2)+2*a*d where vf is the final velocity and d is the distance traveled.
C: vf=vo+a*t
D: d=t*(vf+vo)/2

To find out how high the ball went, you need to use D. But to use D you first need to find the time the ball traveled upward and you must use C to get that.
The initial spee of the ball is 45m/s, but not all of that velocity is going upward. Only the sine of 29 degrees*45m/s is going upward. So use The sine of 45 as the initial velocity (vo).
Then set the final velocity (vf) to be zero, the point at which the ball stops moving upward and is at its max height. You know that the accelleration (a) is (-9.8/3.5)m/s^2 (negative because it is a decelleration). So just solve the equation for t.
Then you can use t in equation D. Again, vo is sin(29)*45m/s and vf is 0. Solve for d. This is the max height in meters.
To find the range, you simply double the time (t) to represent the time the ball goes all the way up and then all the way down, and since we are not using air resistance, it is simply a matter of the speed times the time to get the distance. But keep in mind that the speed is not 45m/s, it is cos(29)*45m/s.
Good luck.
PS: did you know that in the days of Napoleon and the American Revolution, this knowledge was considered "Top Secret"?