At time t = 0 a small ball is projected from point A with a velocity of 292 ft/sec at the angle 56°. Neglect atmospheric resistance and determine the two times t₁ and t2 when the velocity of the ball makes an angle of 24° with the horizontal x-axis. t₁ = u = 292 ft/sec Answers: t2 = 56 i i sec sec
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- 1110. In Fig. P-1109, a ball thrown down the incline strikes it at a distance s 254.5 It. If the ball rises to a maximum height h 64.4 ft above the point of release, compute its initial velocity vo and inclination 6. Ans. v. = 80.5 ft per see; 0= 53.1° FIG. P-1109 and P-1110.sin(20) where vo is the initial velocity in feet per second and is the angle of elevation. If yo 2300 feet per second and is changed from 13° to 14° use differentials to approximate the change in the range. Round your answer to the nearest integer. The range R of a projectile is R = Select one: Ⓒa. 2811 ft X b. 2593 ft C. 5623 ft d. 4568 ft 5186 ft e. = Incorrect 32PROBLEM: A ball is kicked at the edge of a cliff with a magnitude of 20m/ s at angle of 40° from the horiz ontal. It hits a vertical wall 5 Øm away and ation of the cliff. Determine th lower than the elev e vertical height from the point where the ball hits the wall to: a. the level of the cliff b. its highest point
- A soccer ball was kicked from the top of the building 37m height with a velocity of 47 m/s and 55° with the vertical. Air resistance neglected. Determine the following: a) The time of the soccer ball reach the maximum height from the roof. b) The maximum height reached by the soccer ball above the ground. c) the time of the soccer ball to hit the ground from its maximum height. d) The maximum range of the soccer ball from the building to the point where the ball hit the ground.A projectile is launched from and returns to ground level, as the figure shows. There is no air resistance. The horizontal range of the projectile is measured to be R= 177 m, and the horizontal component of the launch velocity is vox = 22.0 m/s. Find the vertical component voy of the projectile. +v voy Var=+22.0m/s to +x Number i R=177 m UnitsThe 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.
- Max Height = V;*sin's (visine) 28 28 Max Ronge = v? sin 20 Vi Height 8 = 9. 81 m/sz Range 1.) A very small cannon (very small, meaning we can disregard its size) has a muzzle velocity (velocity of the projectile as it leaves the barrel) of 30 m/s and is at an angle of 35° from the ground. a.) What is the Max Height? b.) What is the Range? Somple . Diagrom Vi = inital velocty cmuzzle velocity) 0 = ongle from the grond Vi = 30 m/e Height e= 350 Ronge a) Monx Height : (30) (ain 5) Max Height = V;? sin?e 28 15- 09 m R(9.81) b.) Ronge : v.* sin 20 (30)2 sin (70*) Ronge = 86.21M * Incicode your final answers. In this case, I used a box21. The maximum distance s and the maximum height h that a projectile shot at an angle 0 are given by: sin 20 and h = vo sin e 2g where vo is the shooting velocity and g = 9.81 m/s?. Determine s(0) and h(0) for 0 = 15°, 25°, 35°, 45°, 55°, 65°, 75° if vo = 260m/s.An object is launched at a velocity of 20 m/s in a direction making an angle of 45° upward with the horizontal. a) What is the maximum height reached by the object? b) If the object has a time of flight of 2.9 s, what is the horizontal range of the object?
- Question 4: A particle is thrown into air with initial speed vo making angle 0 with the horizontal as shown in Figure 1. The gravitational acceleration g and a constant acceleration a, due to a wind act on the particle. Find the range of the motion. a, 'w Figure 1 100An object is launched at a velocity of 20 m/s in a direction making an angle of 25° upward with the horizontal. a) What is the maximum height reached by the object? b) What is the horizontal range (maximum x above ground) of the object?A projectile is launched off of a 10.0 meter high roof with an initial velocity of 29.0 m/s in a direction that makes a 38.0° angle with thehorizontal.What are the x and y components of its initialvelocity? a)Whatis the object’s maximum height relative to groundlevel? Answers: a) 22.85 m/s and 17.85 m/s