Question #60405

Ques1. A body thrown vertically upward and returns back to the thrower after 6 seconds. Find- 1.velocity with which the body is thrown. 2.distance and
displacement of the body after- (a) 1 sec (b) 2 sec (c) 3 sec (d) 4 sec (e) 5 sec (f) 6 sec 3. Using given data plot- (a) distance time graph (b) displacement time graph (c) speed time graph (d) velocity time graph (e) acceleration time graph 4. Can a body have zero velocity while moving with uniform acceleration. Explain with graph.
1

Expert's answer

2016-06-17T09:59:02-0400

Question #60405, Physics / Other

Ques1. A body thrown vertically upward and returns back to the thrower after 6 seconds. Find- 1. velocity with which the body is thrown. 2. distance and

displacement of the body after- (a) 1 sec (b) 2 sec (c) 3 sec (d) 4 sec (e) 5 sec (f) 6 sec 3. Using given data plot- (a) distance time graph (b) displacement time graph (c) speed time graph (d) velocity time graph (e) acceleration time graph 4. Can a body have zero velocity while moving with uniform acceleration. Explain with graph.

Solution

The vertical position of the body relatively to the thrower:


y=v0tgt22y = v _ {0} t - \frac {g t ^ {2}}{2}


When the body returns to the thrower, t=6t = 6 and y=0y = 0 :


6v018g=0;6 v _ {0} - 1 8 g = 0;v0=3g;v _ {0} = 3 g;v0=29.4m/sv _ {0} = 2 9. 4 \mathrm {m / s}


Plugging the initial velocity into the original equation:


y=29.4t4.9t2y = 2 9. 4 t - 4. 9 t ^ {2}


Plugging tt values and calculating yy :



Distance-time graph:



Displacement-time graph:



The velocity is the first derivative from displacement:


y=29.49.8ty ^ {\prime} = 2 9. 4 - 9. 8 t


Speed is the absolute value of the velocity:


y=29.49.8ty ^ {\prime} = \left| 2 9. 4 - 9. 8 t \right|


The acceleration is the free fall acceleration: a=g=9.8a = g = 9.8


The body has zero acceleration at the highest point (t=3)(t = 3) . At this point, the velocity-time graph crosses the xx -axis.

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