Question

A Bullet is fired with a velocity of 9.8m/s in a direction making an angle of 60degree with the vertical calculate the time of flight and the maximum height reached by it
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Answer 1

Answer:

3

Explanation:

this is the answer for this question

Answer 2

Angle bullet and vertical is 60° given. So angle between horizontal and bullet will be 90°-60°=30°= a (let)

Initial velocity u = 9.8m/s (given)

Time of flight T is given by

T = 2u×sina/g

g= 9.8m/s^2 and sin30° = 1/2

So T = (2×9.8×1)/9.8×2 = 1 second

Maximum height H is given by (u×sina)^2/2g

So H= (9.8×1/2)^2/2×9.8

= 24.01/19.6 = 1.225m