1. A marble rolled off of a 1-m high tabletop with an initial horizontal velocity of 10 m/s.a) Determine its final velocity.b) How much time will it take for the marble to hit the floor?c) How far will the ball head forward before reaching the floor?2. A ball is hurled from a horizontal ground with a speed of 45 m/s at an angle of 35 0 from thehorizontal.a) What is the final horizontal velocity of the ball after 3.5 seconds?b) How much time will it take for it to reach its highest point?c) How much time will it take for it to reach the ground?d) What is the maximum height reached by the ball?e) What is the maximum horizontal distance covered by the ball?
Question
- A marble rolled off of a 1-m high tabletop with an initial horizontal velocity of 10 m/s.a) Determine its final velocity.b) How much time will it take for the marble to hit the floor?c) How far will the ball head forward before reaching the floor?2. A ball is hurled from a horizontal ground with a speed of 45 m/s at an angle of 35 0 from thehorizontal.a) What is the final horizontal velocity of the ball after 3.5 seconds?b) How much time will it take for it to reach its highest point?c) How much time will it take for it to reach the ground?d) What is the maximum height reached by the ball?e) What is the maximum horizontal distance covered by the ball?
Solution
Sure, let's solve these problems one by one.
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a) The final velocity of the marble can be determined using the equation of motion: v = u + gt. Here, u is the initial velocity (0 m/s in the vertical direction), g is the acceleration due to gravity (9.8 m/s²), and t is the time it takes to hit the floor. However, we don't know the time yet, so we need to calculate that first.
b) The time it takes for the marble to hit the floor can be calculated using the equation: h = ut + 0.5gt². Here, h is the height of the tabletop (1 m). Solving for t gives us t = sqrt(2h/g) = sqrt(2*1/9.8) = 0.45 seconds.
c) The horizontal distance the marble travels before hitting the floor can be calculated using the equation: d = vt. Here, v is the horizontal velocity (10 m/s) and t is the time it takes to hit the floor (0.45 seconds). So, d = 10*0.45 = 4.5 meters.
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a) The horizontal velocity of the ball remains constant throughout its flight (assuming no air resistance). So, the final horizontal velocity after 3.5 seconds is still 45 m/s.
b) The time it takes to reach the highest point can be calculated using the equation: u = gt. Here, u is the initial vertical velocity, which can be calculated as u = vsin(θ) = 45sin(35) = 25.7 m/s. Solving for t gives us t = u/g = 25.7/9.8 = 2.62 seconds.
c) The total time of flight can be calculated as 2t (time to reach the highest point), which is 22.62 = 5.24 seconds.
d) The maximum height can be calculated using the equation: h = ut - 0.5gt². Here, u is the initial vertical velocity (25.7 m/s) and t is the time to reach the highest point (2.62 seconds). So, h = 25.72.62 - 0.59.8*2.62² = 34.1 meters.
e) The maximum horizontal distance can be calculated using the equation: d = vt. Here, v is the horizontal velocity (45 m/s) and t is the total time of flight (5.24 seconds). So, d = 45*5.24 = 235.8 meters.
Similar Questions
3. A ball rolls off of a table with a speed of 3.2 m/s. The table is 1.5 m high.a) When does the ball hit the ground? (Answer=0.55 s)b) How far away from the base of the table does the ball travel? (Answer=1.8 m)c) With what speed does the ball hit the floor? (Answer=6.3 m/s)
A small ball rolls horizontally off the edge of a tabletopthat is 1.20 m high. It strikes the floor at a point 1.52 m hori-zontally from the table edge. (a) How long is the ball in the air?(b) What is its speed at the instant it leaves the table?
Every word problem, there should be a conclusion. therefore add a conclusion to the following answer. "(a) The velocity of the ball at its highest point is 0 m/s. This is because at the highest point, the ball stops moving upwards and is about to start falling downwards. (b) The velocity of the ball 1 second before it reaches its highest point depends on the initial speed and the acceleration due to gravity. The acceleration due to gravity is approximately 9.8 m/s². So, if we denote the initial speed as v0, then the velocity 1 second before the highest point is v0 - 9.8 m/s. (c) The change in velocity during this 1-second interval is the final velocity minus the initial velocity, which is (v0 - 9.8 m/s) - v0 = -9.8 m/s. (d) The velocity of the ball 1 second after it reaches its highest point is -9.8 m/s. This is because the ball has started to fall and is accelerating downwards due to gravity. (e) The change in velocity during this 1-second interval is the final velocity minus the initial velocity, which is -9.8 m/s - 0 = -9.8 m/s. (f) The change in velocity during the 2-second interval is the final velocity after 2 seconds minus the initial velocity, which is (-9.8 m/s) - (v0 - 9.8 m/s) = -19.6 m/s. (g) The acceleration of the ball during any of these time intervals and at the moment the ball has zero velocity is -9.8 m/s². This is the acceleration due to gravity, and it is always acting on the ball, regardless of its velocity."
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5. A ball is thrown with enough speed straight up so that it is in the air several seconds. (a) What is the velocity of the ball when it gets to its highest point? (b) What is its velocity 1 s before it reaches its highest point? (c) What is the change in its velocity during this 1-s interval? (d) What is its velocity 1 s after it reaches its highest point? (e) What is the change in velocity during this 1-s interval? (f) What is the change in velocity during the 2-s interval? (Careful!) (g) What is the acceleration of the ball during any of these time intervals and at the moment the ball has zero velocity?
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