In shot putting, many athletes elect to launch the shot at an angle that is smaller than the theoretical one (about 42°) at which the distance of a projected ball at the same speed and height is greatest. One reason has to do with the speed the athlete can give the shot during the acceleration phase of the throw. Assume that a 7.260 kg shot is accelerated along a straight path of length 1.650 m by a constant applied force of magnitude 380.0 N, starting with an initial speed of 2.500 m/s (due to the athlete’s preliminary motion). What is the shot’s speed at the end of the acceleration phase if the angle between the path and the horizontal is (a) 30.00° and (b) 42.00°? (Hint: Treat the motion as though it were along a ramp at the given angle.) (c) By what percent is the launch speed decreased if the athlete increases the angle from 30.00° to 42.00°?
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Fundamentals Of Physics - Volume 1 Only
- A shot putter launches a 7.260 kg shot by pushing it along a straight line of length 1.650 m and at an angle of 34.10° from the horizontal, accelerating the shot to the launch speed from its initial speed of 2.500 m/s (which is due to the athlete’s preliminary motion). The shot leaves the hand at a height of 2.110 m and at an angle of 34.10 degrees, and it lands at a horizontal distance of 15.90 m. What is the magnitude of the athlete’s average force on the shot during the acceleration phase? (Hint: Treat the motion during the acceleration phase as though it were along a ramp at the given angle.)arrow_forwardAn industrious chipmunk creates an acorn catapult. This acorn catapult fires a 0.00543 kg acorn at an initial speed of 21.8 m/s. The catapult can be moved to any horizontal distance from the target and it can be set to shoot the acorn at any angle between 14° and 86.6°. The chipmunk will adjust the distance and angle until he gets a shot just right, and he deems a shot to be successful when the target is hit at the moment the acorn reaches its apex d) If the acorn hits the target and bounces back, losing 24.2% of its speed in the collision, what is the magnitude of the change in the acorn's momentum from immediately before the collision to immediately after the collision? e) If the duration of the collision described in part D is 3.01 x 10^-5 s, what is the magnitude of the average force the acorn imparts onto the target?arrow_forwardA rock is thrown off a cliff at an angle of 54° above the horizontal. The cliff is 110 m high. The initial speed of the rock is 20 m/s. (Assume the height of the thrower is negligible.)arrow_forward
- A tennis ball is thrown with an initial velocity of 15.0 m/s at an angle of 60.0° above the horizontal. If air resistance and wind are neglected, the maximum height that the ball can reach is 43.3 m 8.61 m 6.81 m 12.8 marrow_forwardA 121 kg crate is sitting at the top of a ramp, which is inclined at an angle of 18 degrees with respect to the horizontal. The crate is a height 0.51 m from the ground, measured vertically from the ground to the crate. Someone gives the crate a quick shove to get it moving, after which it slides down the ramp without any further assistance. The coefficient of kinetic friction between the crate and the ramp is Hs = 0.23. How much time (in seconds) does it take for the crate to get to the %3D base of the ramp?arrow_forwardA football is thrown from an initial height of 1.95 m above the ground. The football's initial velocity components are Vi = 6.10 m/s and v₁ = 5.30 m/s. XI Determine the football's maximum height above the ground. Ignore air resistance and any other nonconservative forces. marrow_forward
- A toad with mass of 280 g leaps into the air to catch a fly. If the toad's jump was at an angle of 35 degrees with respect to the horizontal, and with an initial velocity of 5.1 m/s, what is its speed (in m/s) when it reaches its maximum vertical displacement?arrow_forwardAt t = 0.00 s a ball is started rolling up an inclined plane with an initial velocity of 6.00 m/s, 15.0°. At t = 2.00 s the ball reverses its direction and begins to roll back down. (a) How far up the slope does the ball travel? (b) Find the ball's acceleration. (c) Find the speed of the ball at t = 3.00 s. (d) Find the distance traveled by the ball during these 3.00 seconds. (e) Find the ball's position at t = 3.00 s.arrow_forwardThe shot put is a track-and-field event in which athletes throw a heavy ball - the shot - as far as possible. The best athletes can throw the shot as far as 23 m . Athletes who use the "glide" technique push the shot outward in a reasonably straight line, accelerating it over a distance of about 2.0 m . What acceleration do they provide to the shot as they push on it? Assume that the shot is launched at an angle of 37 degrees, a reasonable value for an excellent throw. You can assume that the shot lands at the same height from which it is thrown; this simplifies the calculation considerably, and makes only a small difference in the final result.arrow_forward
- A block with mass m = 14.9 kg slides down an inclined plane of slope angle 18.8o with a constant velocity. It is then projected up the same plane with an initial speed 2.95 m/s. How far up the incline will the block move before coming to rest?arrow_forwardA 4 kg squirrel is launched from a cannon 7m above sea level and is fired at a 45 degree angle from the ground, launching the squirrel at 150 m/s. At the squirrel's peak height, an eagle snatches the squirrel and takes a leave of 303 degrees. Assume that the additional weight of the squirrel cuts the air speed velocity of the eagle by 75%. After flying for two minutes, the squirrel drops an acorn. A wing with a heading of 155 degrees accelerates the acorn at 7 m/s^2. Where does the acorn land? Assume that: 1. The squirrel reaches an instant velocity upon firing the canon. 2. Gravitational acceleration is 10m/s^2. 3. Air resistance is negligible. 4. The eagle experiences no knockback and cancels out the squirrel's velocity. 5. The air speed velocity of the eagle is 12 m/s. 6. The acorn sat around so long that it is now deep underground.arrow_forwardA student launches a small rocket which starts from rest at ground level. At a height h=1.04km, the rocket reaches a speed of vf=391m/s. At that height, the rocket runs out of fuel, so there is no longer any thrust propelling it. Take the positive direction to be upward. After the rocket's engine turns off at a height of h=1.04kmℎ=1.04km, it continues to move upward due to the velocity that it reached. Calculate the maximum height, in meters above ground level, that the rocket reaches.arrow_forward
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