Considered Formula; Motion Under Gravity
i)v=u+gtExample 1
A stone of mass 2kg is released from a height of 20m above the ground. Find
a)
Velocity reached just before hit the ground
b) Time taken to reach the ground
Solution
Data given
Distance (s) = 20 m
Accelaration due to gravity (g) = 10 m/s2
a)
Final velocity (v) = ?
v2=2gs⇒Initial velocity (u)=0
The final velocity = 20 m/s
b)
Time (t) = ?
v=gt
The time taken to reach the ground = 2seconds
Example 2
A stone is thrown vertically upward from the ground with a velocity of 30 m/s. find
a)
Maximum height reached
b) Time taken to reach the maximum height
c) Time taken to reach the ground again
d) The velocity reached half-way to the maximum height
Solution
Data given
Initial velocity (u) = 30 m/s
Final velocity (v) = 0 m/s
Accelaration due to gravity (g) = 10 m/s2
a)
Distance (s) = ?
For upward motion, g = -10 m/s2
v2=u2+2gs
The maximum height reached = 45 m
b)
Time taken to reach maximum height (t1) = ?
v=u+at
The time taken to reach maximum height = 3 seconds
c)
Time taken to reach ground again (t2) = ?
t2=2t1
The time taken to reach ground again = 6 seconds
d)
Final velocity (v) = ?
Time (t)=(32)=1.5 sec
v=u+at
The velocity reached half-way to the maximum height = 15 m/s
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