A basketball star covers 2.80 m horizontally in a jump to dunk the ball (Fig. P4.12a). His motion through space can be modeled precisely as that of a particle at his center of mass , which we will define in Chapter 9. His center of mass is at elevation 1.02 m when he leaves the floor. It reaches a maximum height of 1.85 m above the floor and is at elevation 0.900 m when he touches down again. Determine (a) his time of flight (his “hang time”), (b) his horizontal and (c) vertical velocity components at the instant of takeoff, and (d) his takeoff angle. (e) For comparison, determine the hang time of a whitetail deer making a jump (Fig. P4.12b) with center of mass elevations y i = 1.20 m, y max = 2.50 m, and y f = 0.700 m. Figure P4.12
A basketball star covers 2.80 m horizontally in a jump to dunk the ball (Fig. P4.12a). His motion through space can be modeled precisely as that of a particle at his center of mass , which we will define in Chapter 9. His center of mass is at elevation 1.02 m when he leaves the floor. It reaches a maximum height of 1.85 m above the floor and is at elevation 0.900 m when he touches down again. Determine (a) his time of flight (his “hang time”), (b) his horizontal and (c) vertical velocity components at the instant of takeoff, and (d) his takeoff angle. (e) For comparison, determine the hang time of a whitetail deer making a jump (Fig. P4.12b) with center of mass elevations y i = 1.20 m, y max = 2.50 m, and y f = 0.700 m. Figure P4.12
Solution Summary: The author calculates the upward motion of the basketball star's flight from the moment it leaves the floor until just before he lands.
A basketball star covers 2.80 m horizontally in a jump to dunk the ball (Fig. P4.12a). His motion through space can be modeled precisely as that of a particle at his center of mass, which we will define in Chapter 9. His center of mass is at elevation 1.02 m when he leaves the floor. It reaches a maximum height of 1.85 m above the floor and is at elevation 0.900 m when he touches down again. Determine (a) his time of flight (his “hang time”), (b) his horizontal and (c) vertical velocity components at the instant of takeoff, and (d) his takeoff angle. (e) For comparison, determine the hang time of a whitetail deer making a jump (Fig. P4.12b) with center of mass elevations yi = 1.20 m, ymax = 2.50 m, and yf = 0.700 m.
A cannon ball is fired with an initial speed of 123 m/s at angle of 60 degrees from the horizontal.
Express the initial velocity as a linear combination of its unit vector components.
( m/s) + ( m/s)
At the maximum height, the speed of the cannon ball is v = m/s and the magnitude of its acceleration is a = m/s2.
The time needed to reach maximum height is t = s.
The maximum height reached by the cannon ball is H = m.
A basketball player who is 2.00 m tall is standing on the floor 10.0 m from the basket as in figure. If he shoots the ball at a 40.0° angle with the horizontal at what initial speed must he throw so that it goes through the hoop without striking the backboard? the basket height is 3.05 m.
Irina, standing on the edge of a cliff, throws a stone at a 40-degree angle above the horizontal with an initial speed of 10.0 m/s. The instant before the stone hits the ground below, it is traveling at a speed of 54 m/s. If Irina were to throw the other rock at a 40-degree angle below the horizontal from the cliff instead, with the same initial speed of 10.0 m/s, what is the magnitude of the velocity of the stone just before it hits the ground? Ignore any effects of air resistance.
Chapter 4 Solutions
Physics for Scientists and Engineers, Technology Update (No access codes included)
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