A bullet has a speed of 2000 ft/s as it leaves a rifle. If it fired horizontally from a cliff 24 ft above a lake, find the range of the bullet.
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A: Here we use equation of motion to solve the problem
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- In the absence of air resistance, a projectile is launched from and returns to ground level. It has a range of 16m. Suppose the launch speed is doubled, and the projectile is fired at the same angle above the ground. What is the new range?A projectile is fired from ground level with an initial speed of 450 m/sec and an angle of elevation of 30 degrees. Use that the acceleration due to gravity is 9.8 m/sec². The range of the projectile is 0 meters. The maximum height of the projectile is meters. The speed with which the projectile hits the ground is m/sec.A projectile is launched with a speed of 140 m/s at an angle of 45 degrees above the horizontal, from a height of 1500 m above the ground. How far does it travel in the horizontal direction before hitting the ground?
- The citizens of Paris were terrified during World War I when they were suddenly bombarded with shells fired from a long-range gun known as Big Bertha. The barrel of the gun was 36.6 m long, and it had a muzzle speed of 2.20 km/s. When the gun’s angle of elevation was set to 55.0°, what would be the range? For the purposes of solving this problem, ignore air resistance. (The actual range at this elevation was 121 km; air resistance cannot be ignored for the high muzzle speed of the shells.) kmA projectile launched at 45° upwards from the ground takes 2.4 seconds to return to the ground. Approximately what speed was the projectile launched?A cannon ball is fired with a velocity of 300 m/s at an angle of 10° with respect to the ground. How long will the cannonball fly through the air? (Ignore the height of the cannon and assume the cannon ball is fired from ground level.)
- Consider a projectile launched at a height h feet above the ground and at an angle ? with the horizontal. If the initial velocity is v0 feet per second, the path of the projectile is modeled by the parametric equations x = t(v0 cos(?)) and y = h + (v0 sin ?)t - 16t2.Let h=6, v0=63.25, and ?=0.79. What is the range of the projectile (i.e., how far does it travel? Round your answers to two decimal places).The initial speed of a tennis ball is 57.5 m/s and the launch angle is ?i = 16°. Neglect air resistance.Question 1) What is the maximum height, h, of the tennis ball? in mQuestione 2) What is the range, R, of the tennis ball? in m info given: what angle results in the greatest height=90 For this 90° angle, the horizontal component of velocity is zero, so the initial velocity is completely in the vertical direction, resulting in greatest height Kato tries substituting ty,max for t, 0 for yi, and h for yf, and gets h = vi2 sin2 ?i 2g When ?i = 90°, sin2 ?i is maximum, so h is maximum what angle results in the maximum horizontal range=45° I substituted 2vi sin ?i g into the expression for the horizontal component of velocity, (vi cos ?i)t, and got R = 2vi2 sin ?i cos ?i g when ?i = 45°, 2?i = 90°, and sin 90° = 1, which is its maximum value.A tennis ball is projected at ground level. After 2 seconds, it just clears a net 2.5 metres high at a horizontal distance of 30 metres. Find:i) The angle of projection and the initial velocity ii) The time of flight and the range
- A golf ball was hit horizontally with a horizontal velocity of 40.0m/s from a height of 120m. How long does it take the ball to hit the ground?Your bow can fire arrows at 70 ??. You shoot an arrow straight up into the air. After 1 second, you shoot anotherarrow into the air.a) If you are shooting the arrows from 2 m above the ground, how far above the ground do the arrowscollide? (Answer: 251 m)