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The velocity of a ball thrown upward at W ft/sec will gradually decrease due to gravity until it reaches its highest point (at which the velocity will be 0 ft/sec). After reaching the peak, the ball will then start to fall back down, increasing in velocity until it reaches the ground.
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What is the velocity of a ball thrown upward at 16 feetsec?
The velocity of the ball is 16 feet/sec when it is thrown upward.
What is the velocity of a ball thrown upward at 16 ft-sec?
The velocity of a ball thrown upward at 16 ft/sec would be 16 ft/sec when it leaves the hand, but it will decrease due to gravity as it moves upward.
What is tje velocity of a ball thrown upward at 16 ft/sec?
If a ball is thrown vertically upward with a velocity of 160 ft/s, then its height after t seconds is s = 160t - 16t^2. If a ball is thrown vertically upward with a velocity of 160 ft/s, then its height after t seconds is s = 160t − 16t^2.
What is the velocity of a ball thrown at 16 ft sec?
The velocity of a ball thrown at 16 ft/sec is 16 ft/sec. This means that the ball is moving at a constant speed of 16 feet per second in a particular direction.
How high can a ball go?
The maximum height of a thrown ball is dependent on the upward portion of the initial velocity. Air friction will slow it somewhat but gravity will cause it to lose most of its upward velocity. The velocity will decrease by 9.8 m/sec for each second of its upward motion, until it reaches zero. At that point, the ball is pulled back toward Earth.
What is the velocity of a ball thrown upward at 16 feetsec?
The velocity of the ball is 16 feet/sec when it is thrown upward.
What is the velocity of a ball thrown upward at 16 ft-sec?
The velocity of a ball thrown upward at 16 ft/sec would be 16 ft/sec when it leaves the hand, but it will decrease due to gravity as it moves upward.
What is tje velocity of a ball thrown upward at 16 ft/sec?
If a ball is thrown vertically upward with a velocity of 160 ft/s, then its height after t seconds is s = 160t - 16t^2. If a ball is thrown vertically upward with a velocity of 160 ft/s, then its height after t seconds is s = 160t − 16t^2.
What is the velocity of a ball thrown at 16 ft sec?
The velocity of a ball thrown at 16 ft/sec is 16 ft/sec. This means that the ball is moving at a constant speed of 16 feet per second in a particular direction.
When a ball is thrown vertically upward at initial velocity of 15 meter per second Find the position and velocity after 1.0 second and 40 second after leaving your hand?
1 sec : position = 10.1 metres above your hand, velocity = 5.2 ms^-1.40 sec : position = 7240 metres below your hand, velocity = 377 ms^-1 downwards.
How high can a ball go?
The maximum height of a thrown ball is dependent on the upward portion of the initial velocity. Air friction will slow it somewhat but gravity will cause it to lose most of its upward velocity. The velocity will decrease by 9.8 m/sec for each second of its upward motion, until it reaches zero. At that point, the ball is pulled back toward Earth.
The velocity of a baseball 4 seconds after it is thrown vertically upward with a speed of 32.1 meter per sec is?
V = V0 + a t V0 = + 32.1 m/sec a = - 9.78 m/sec2 t = 4 sec V = (32.1) + 4 (-9.78) = (32.1) - (39.12) = - 7.02 m/sec (7.02 m/sec downward)
You throw a ball upward with an initial speed of 20 m sec About 4 seconds later the ball returns and you catch it How fast was the ball traveling downward when you caught it?
20m/sec
How does gravity affect how high a ball bounces?
Gravity is the force opposing the upward velocity of the ball, which has mass m The ball bounces at velocity v, so kinetic energy = 1/2m x v squared If the ball rises height h, energy to do this is m x g x h These two terms are equal, so solve for h. Using SI units, g = 9.81 meters/sec/sec, m in kg, h in meters, v in meters/sec This ignores air resistance
When throwing a ball straight up when in the air with an initial velocity of 10 meters per second what high will it go and how long will it take to return to the ground?
The initial velocity is 10 meters/sec and is thrown up against the gravitational pull of the earth. This means that the ball is experiencing a deceleration at the rate of 9.8 meters/sec/sec to bring its final velocity to zero. v^2 - u^2 = 2gs where u is the initial velocity, v the final velocity, g is the acceleration or deceleration, and s is the distance traveled. 0^2 - 10^2 = 2 x (-9.8) x s -100 = -19.6s 100 = 19.6s s = 100/19.6 = 5.102 meters Now v = u + gt where v is the final velocity, u is the initial velocty, g is the acceleration or deceleration, and t is the time. When the ball is thrown up with 10 meters/sec velocity it is acted upon by the deceleration of gravity until its velocity becomes zero. So 0 = 10 - 9.8t or 9.8t = 10 t = 1.020 seconds The time for the ball to go up is 1.020 seconds and the same time is taken for the ball to come back for a total of 2.040 seconds.
A ball is thrown vertically upward. What are its velocity and acceleration when it reaches its maximum altitude What is the acceleration of the ball just before it hits the ground?
At the maximum height the ball will be completely stopped from moving upward or downward; thus the speed of the ball would be 0 mph. The ball is only stopped for a split second and then it begins moving downward, then increasing at 9.81m/s^2 until it reaches maximum velocity.
What shape does the path of a thrown ball follow?
The ball follows a parabolic path when thrown. In a vacuum (with no air or other forces acting upon it) the gravitational pull of the earth causes the ball to accelerate toward the earth (9.8m/sec
