A bullet fired vertically up from the ground reaches a height 40m in i...
Understanding the problem:
A bullet is fired vertically up from the ground and reaches a height of 40m in its path. After reaching this point, it takes further 2 seconds to come back to the ground. We need to find out the total time for which the bullet remains in the air.
Solution:
To solve this problem, we need to consider the motion of the bullet in two parts:
1. The time it takes to reach the maximum height of 40m.
2. The time it takes to come back to the ground.
Calculating time to reach the maximum height:
To calculate the time taken by the bullet to reach the maximum height, we can use the formula:
v^2 - u^2 = 2as
where,
v = final velocity (0 m/s as the bullet reaches the maximum height)
u = initial velocity (unknown)
a = acceleration due to gravity (-9.8 m/s^2)
s = displacement (40m)
Substituting the values, we get:
0^2 - u^2 = 2(-9.8)(40)
u^2 = 2(-9.8)(40)
u = 39.2 m/s
Now, we can use the formula:
v = u + at
where,
v = final velocity (0 m/s as the bullet reaches the maximum height)
u = initial velocity (39.2 m/s)
a = acceleration due to gravity (-9.8 m/s^2)
t = time taken to reach the maximum height
Substituting the values, we get:
0 = 39.2 - 9.8t
t = 4 seconds
Calculating time to come back to the ground:
Since the bullet takes 2 seconds to come back to the ground, we can directly use this value.
Total time for which the bullet remains in the air:
The total time for which the bullet remains in the air is given by:
Total time = time to reach maximum height + time to come back to the ground
Total time = 4 + 2 = 6 seconds
Therefore, the bullet remains in the air for 6 seconds.
A bullet fired vertically up from the ground reaches a height 40m in i...
40=1/2a(t+2)^2
To make sure you are not studying endlessly, EduRev has designed Class 11 study material, with Structured Courses, Videos, & Test Series. Plus get personalized analysis, doubt solving and improvement plans to achieve a great score in Class 11.