physics projectile motion problems help please?

Question

i have two problems i cant get please help

A) A bullet is shot at a target 200 ft away. if the bullet is shot at a height of 5 feet, with an initial velocity of 150 ft/s, and at an angel of 10 with the horizontal, when will it reach the target and what will be the height where it hits?

B) a disk is thrown from a height of 5 meter at an initial velocity of 65 m/s at an angle of 10 with the ground. after .5 second, how far has the disk traveled horizontally and vertically?

please help ireally dont understand

Answer 1

Bullets are projectiles, and like other projectiles, do NOT move in a straight line. They follow a parabolic path just like everything else... only much faster!

a) since everything is in the English system, we need to know g = 32.2 ft/s² downward.

Break things up into x and y components..

x-direction (left and right)
d = 200 ft
vxi = 150 * cos(10º) ft/s = 147.72 ft/s
a = 0 ft/s²

d is the distance traveled
vxi is the initial velocity in the x direction
a is the acceleration in the x direction

Since a = 0, we know that

d = vix * t

200 = 147.72t
t = 1.35 s

In the y-direction
h = 5 ft
g = -32.2 ft/s²
vyi = 150 * sin(10º) ft/s = 26.05 ft/s
y = final height

h = initial height
g = acceleration due to gravity
vyi = initial velocity in the y-direction

y = ½g t² + vyi t + h
y = .5 * -32.2 * 1.35² + 26.05 * 1.35 + 5
y = 10.75 ft above the ground

B) This is the same idea as earlier, except g = 9.8 m/s² since we're now in metric units.

In the x-direction....
vx = 65 * cos(10º) m/s = 64.01 m/s
t = 0.5 s

d = vx * t = 65.01 * 0.5 = 32 m in the x direction

In the y-directoin
h = 5 m
vyi = 65 * sin(10º) m/s = 11.29 m/s
g = -9.8 m/s²
t = 0.5 s

y = ½ g t² + vyi t + h
y = .5 * -9.8 * .5² + 11.29 * .5 + 10
y = 14.42 m

So the disk will have moved 32 m over and 14.42 m up.

Answer 2

A) The time it takes to reach the target depends only on the horizontal component of velocity. That is v0*cosø, where ø is the angle with the horizontal. The distance to target is d0, so the time t = d0/[v0*cosø] The vertical component of velocity is v0*sinø, and the vertical trajectory is defined by h = h0 + vy*t - .5*g*t^2, where vy is the vertical component of initial velocity. Use the time from the first calc in the second formula to get h at the target. g = gravitational acceleration.

B) The same formulas are used.

Answer 3

You do not have enough info to answer A. The caliber (cross section) of the bullet will determine atmospheric drag along with altitude. Are you shooting at sea level or maybe in the mountains?? A good snipers job is never done.
B)horizontal is .5x65(sin10)
vertical is .5x65(cos10)

Answer 4

a million. First we would desire to stumble on how long the cannonball is interior the air, its "draw close time" in case you will. This span of time is the time it takes for the cannonball to pass up, and fall back off. it is thoroughly based on the y-portion of the rate, so we ought to continuously use some trigonometry to stumble on this. in case you drew a suitable triangle, you will locate that the y-comp onent: velocity at y axis = (preliminary velocity)sin(seventy 9) = 993sin79 = 974.756 m/s. 2. Now which you have the rate on the y axis, we can now locate the time it takes for the cannon to land. we can use the equation very final velocity = preliminary velocity + (Acceleration)(Time) or Vf = Vi + at be conscious that your aceeleration = -g, using fact it fairly is pulling it downward so: Vf = Vi - gt. putting apart t: Vf + gt = Vi gt = Vi - Vf t = (Vi - Vf)/g all of us comprehend the preliminary velocity interior the y axis, that's 974.756 m/s. yet what concerning to the appropriate velocity? right here is a few thing exciting: once you throw a ball up at say a definite velocity 2m/s, whilst it falls back on your hand (assuming you saved your hand on a similar place) that's going to return on your hand at a velocity of -2m/s, the different of the preliminary velocity. So contained regarding the cannonball, i'm able to declare that the appropriate velocity is -974.756 m/s. i comprehend my Vi, Vf, and g, so i'm able to now resolve for time. t = (974.756 - (-974.756))/9.8 = (974.756 + 974.756)/9.8 = 1949.512/9.8 = 198.9298 seconds. 3. Why do i'd desire to comprehend the draw close time? that's using fact the ball is vacationing on the x path for this plenty time, and after that factor, it stops shifting (till it keeps on rolling, yet it somewhat is yet another concern! permit's assume it continues to be positioned the place it lands.) besides, i'm able to use the equation velocity = distance/time or distance = (velocity)(time) = vt I even have my time, yet i choose my velocity. remember we are looking for for the area on the x-path, no longer the y, so we for the fee right here, we would desire to get the x-portion of the projectile's velocity. employing trigonometry back: velocity at x axis = (preliminary velocity)cos(seventy 9) = 993cos79 = 993(0.19) = 189.473 m/s. So, employing this on the previous equation, and the air-time: distance traveled alongside the horizontal path = vt = (189.473)(198.9298) = 37,691.826 meters.

Answer 5

A) Ignoring rifling and drag,
t = d/(Vcosφ)
t = 200/(150cos10)
t = 1.3539 s

h = h0 + Vsinφt - (1/2)at^2
h = 5 + 150(sin10)1.3539 - (1/2)(32.2)1.3539^2
h = 10.753 ft

B)h = 65sin10/(2*9.8) = 575.87 cm
d = 2(65)(cos10)(65)(sin(10)/9.8
d = 147.45 m