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Question

A projectile is thrown with a velocity of 102 m/s at an angle of 45 with horizontal. The interval between the moments when speed is 125 m/s is
(g=10 m/s2)

A
1 s
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B
1.5 s
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C
2 s
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D
0.5 s
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Solution

The correct option is A 1 s
ux=102cos45=10 m/suy=102sin45=10 m/s
Let α be the angle with the horizontal when speed is =125 m/s ;
As the horizontal component of velocity always remains constant
102cos45=125cosαcosα=25
i.e.vy=125×sinα=55×15=5 m/s

Using first equation of motion: vy=uyg(t1)5=1010t1t1=12 s
During downward journey: vy=uyg(t2)5=1010t2t2=32 s
t2t1=1 s
The interval between the moments when speed is 125 m/s is 1 s.

Alternative solution

Horizontal component will remain constant
102cos45=125cosαcosα=25sinα=15
For distance AB in path of projectile,
Time period T=2vsinαg=2125×1510=1 s

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