11) A projectile is fired with an initial velocity of 120.0 m/s at an angle, θ, above the horizontal. If the projectile's initial horizontal speed is 55 meters per second, then angle θ measures approximately (A) 13° (B) 27° (C) 63° (D) 75° 12) A golf ball is hit at an angle of 45° above the horizontal.

Solution: The velocity components V x V x and V y V y are given by the formula: In the given problem, V 0 =40m/s V 0 = 40 m / s, ? = 50° and g is 9.8m2 9.8 m 2. The height of the projectile is given by the component y, and it reaches its maximum value when the component V y V y is equal to zero.A projectile is fired from the surface of the earth with a velocity of 5 \[m{{s}^{-1}}\,\] and angle \[\theta \] with the horizontal. Another projectile fired from another planet with a velocity of 3 \[m{{s}^{-1}}\,\]at the same angle follows a trajectory which is identical with the trajectory of the projectile fired from the earth.

The Range R of a Projectile: Another kind of projectile problem is picture below. The ground is flat throughout the trajectory of the projectile. Also suppose 1. the magnitude of the initial velocity is V 0 = 35mês and 2. the direction is initially an angle q=30° above the horizontal direction. (Obviously, these numerical values can be ...If you fire a projectile at an angle, you can use physics to calculate how far it will travel. When you calculate projectile motion, you need to separate out the horizontal and vertical components of the motion. This is because the force of gravity only acts on the projectile in the vertical direction, and the […]A projectile is launched with an initial speed of 60.0 m/s at an angle of 30.00 above the horizontal. The projectile lands on a hillside 4.00s later. Neglect air friction. (a) What is the projectile's velocity at the highest point of its trajectory? (b) What is the displacement of the rocket? 5.Projectile Super Problem A golf ball is hit from the ground at 35 m/s at an angle of 55º. The ground is level. ... to reach the highest point. The height at this time is 2 2 1 2 2 1 2 2 1 sin sin sin ... some angle , sin[2(45 )] sin(90$ 2 ) cos2 ...However, if you throw it at a 45 degree angle, the horizontal velocity becomes 60 2 / 2 mph, or about 42. football at an angle of 40 degrees above the horizontal with an initial speed of 22 m/s. 5 m away from a tower 22 m high. Place the steel ball into the launcher and use the push rod to load the ball until the third. 318 m Range =2.Projectile motion is the motion of an object thrown or projected into the air, subject to only the acceleration of gravity. The object is called a projectile, and its path is called its trajectory.The motion of falling objects, as covered in Chapter 2.6 Problem-Solving Basics for One-Dimensional Kinematics, is a simple one-dimensional type of projectile motion in which there is no horizontal ...Nov 01, 2010 · (B) 140 m (D) 45 m Base your answers to questions 12 and 13 on the information and diagram below. A child kicks a ball with an initial velocity of 8.5 meters per second at an angle of 35º with the horizontal, as shown. The ball has an initial vertical velocity of 4.9 meters per second and a total time of flight of 1.0 second. [Neglect air ... 4.47 A projectile is fired up an incline (with incline angle ) with an initial seed v i, at an angle i, with respect to the horizontal ( i > ) as shown in Figure P4.47. (a) Show that the projectile travels a distance d up the incline, where . d = [ 2 v i 2 cos i sin ( i - ) ] / [g cos 2]Now the total time that the projectile spends moving forward (before it hits the ground) is. Δ t = 9.749 + 9.195 = 18.944 s. Finally, as always, we use the horizontal velocity and the total time to calculate the distance traveled: d = v x ⋅ Δ t = 63.09 m s ⋅ 19.944 s = 1258 m. That's pretty far.Vector and Projectile Motion : MCQ you shouldn’t miss physics - mcq / By Mohit Poudel Scroll down and click “ START TEST ” to begin your model entrance exam.