A projectile is launched with an initial speed of 75m/s at an angle of 40' above the horizon. The projectile lands on a hillside 5.0s later. Neglect air friction. a) What is the projectile's velocity at it's highest point of its trajectory? b) What is the straight-line distance from where the projectile was launched to where it hits its target? (Note that the hill may slope up or down from the launch point.)
I was able to calculate a) as 57.45 m/s, but I am unable to figure out b. Can anyone help me?
2007-09-19 19:27:19 · 4 answers · asked by lostinphysics 2 in Science & Mathematics ➔ Physics
a) As we are neglecting air friction, the horizontal velocity goes undiminished throughout the flight. Therefore vh = V cos(40), where V = 75 mps is the launch pad velocity and, at the apex, the vertical component vv = 0. Therefore the projectile's velocity v = sqrt(vh^2 + vv^2) = vh = 75 cos(40).
b) The horizontal velocity vh = constant = 75 cos(40); so the distance traveled horizontally is H = vh t = 75 cos(40) 5; where t = 5 sec, the time traveled.
You can do the math. The point of this problem is that vh, without drag friction, stays at whatever the launch horizontal velocity was. That results because only weight (the force of gravity) acts on the projectile when there is no air friction. And the force of gravity acts only in the vertical (not horizontal) direction.
2007-09-19 19:59:24 · answer #1 · answered by oldprof 7 · 2⤊ 0⤋
Without doing all the work for you, try d=v sub 0 times t + one half a times t squared to calculate the height of the impact point, and then use Pythagoras for the straight line distance.
By the way, air resistance is stronger than gravity by a substantial degree (precisely how much depends strongly on projectile shape and velocity) so the type of problem that says to neglect air resistance is not a real world problem.
2007-09-19 20:02:11 · answer #2 · answered by George 2 · 0⤊ 0⤋
If the piece separates gently (and not with an explosion as in case of a hearth cracker rocket ) then it somewhat is going to fly component by ability of component as no exterior tension has acted on the stone or the piece. So the the terrific option answer is selection a million. In case of an explosion halfway it somewhat is going to be counted near to separation. The small piece can pass in any random direction (no longer inevitably horizontal) and stick with distinctive parabolic direction.
2017-01-02 10:24:46 · answer #3 · answered by Anonymous · 0⤊ 0⤋
the horizontal velocity is unaffected by acceleration due to gravity. then no matter where it lands the range will be
R= horizontal velocity x time taken
= 75cos40 x 5=375cos40
2007-09-19 21:56:53 · answer #4 · answered by raj 2 · 0⤊ 0⤋