A mango fruit drops to the ground from the top of its tree which is 5m high.how long does it take to rech the gound?
h = -4.9t^2 + 5
we want h=0
4.9t^2 = 5
t^2 = 5/4.9
t = √(5/4.9) = appr 1.01 seconds
i want to know to solve a problem
Niçe
To find the time it takes for the mango fruit to reach the ground, we can use the equation of motion for uniformly accelerated motion:
h = (1/2) * g * t^2
where:
h is the height of the tree (5m),
g is the acceleration due to gravity (approximately 9.8 m/s^2),
t is the time it takes for the mango to reach the ground (unknown).
We need to rearrange this equation to solve for t:
t^2 = (2 * h) / g
Now, substituting the given values:
t^2 = (2 * 5) / 9.8
t^2 = 10 / 9.8
t^2 ≈ 1.02
Taking the square root of both sides, we find:
t ≈ √1.02
t ≈ 1.01 seconds
Therefore, it takes approximately 1.01 seconds for the mango fruit to reach the ground.
Niçe
S=Ut+½at²