an arrow shot vertically upwards loses its intial speed by 60% in 3 secs maximum height reched by the arrow is (g=9.8 m/sec^2)
To find the maximum height reached by the arrow, we first need to find the time it takes for the arrow to reach its maximum height.
Given:
Initial speed (u) = 100% (let's assume the initial speed is 100% for simplicity)
Time taken to lose 60% of initial speed (t) = 3 seconds
Since the arrow loses 60% of its initial speed in 3 seconds, we can calculate the final speed (v) after 3 seconds using the formula:
v = u - (0.6 * u) (where u is the initial speed)
v = 0.4 * u
Now, we know that the final speed when the arrow reaches its maximum height is zero because it momentarily stops before falling back down. So, we can set the final speed to zero and solve for time (t) to reach the maximum height:
0 = v - (g * t) (where g is the acceleration due to gravity, and t is the time taken to reach the maximum height)
0 = 0.4 * u - (g * t)
0.4 * u = (9.8 * t) (substituted for g)
t = (0.4 * u) / 9.8
Since we assumed the initial speed (u) to be 100%, we can substitute u = 1 in the above equation:
t = (0.4 * 1) / 9.8
t ≈ 0.0408 seconds
Now that we know the time taken to reach the maximum height is approximately 0.0408 seconds, we can calculate the maximum height (h) using the formula:
h = (u * t) - (0.5 * g * t^2)
Assuming u = 1:
h = (1 * 0.0408) - (0.5 * 9.8 * 0.0408^2)
h = 0.0408 - (0.5 * 9.8 * 0.001664)
h ≈ 0.0408 - 0.081544
h ≈ -0.040744
The negative value of h indicates that the arrow does not reach a maximum height but falls back down immediately. Therefore, the maximum height reached by the arrow is approximately 0 meters.
The answer with second equation of motion is bringing me the answer same as your all
I mean not bringing
0.4Vo = Vo + g*t.
0.4Vo = Vo + (-9.8*3).
0.6Vo = 29.4.
Vo = 49 m/s.
Vf^2 = Vo^2 + 2g*h.
Vf = 0.
g = -9.8 m/s^2.
h = ?