A palm fruit drops to the ground from the top of a tree 5m tall. How long does it take to reach the ground.
5t^2 = 5
a = -g = -10 m/s^2
v = Vi + a t = 0 -10 t m/s
h = Hi + Vi t + (1/2) a t^2 = 0 + 0 - (10/2) t^2 meters
h = 5 - 5 t^2
when h = 0
t^2 = 5/5 = 1
t = 1 second
To calculate how long it takes for the palm fruit to reach the ground, we can use the equation of motion related to free fall. The equation is as follows:
h = (1/2) * g * t^2
where h is the vertical height, g is the acceleration due to gravity (approximately 9.8 m/s^2), and t is the time taken.
In this case, the initial height (h) of the palm fruit is 5m. Let's solve the equation for t:
5 = (1/2) * 9.8 * t^2
Rearranging the equation, we have:
t^2 = 5 * 2 / 9.8
t^2 = 1.02
Taking the square root of both sides, we get:
t = √1.02
t ≈ 1.01 seconds
Therefore, it takes approximately 1.01 seconds for the palm fruit to reach the ground.
To find the time it takes for the palm fruit to reach the ground, we can use the equation of motion for a freely falling object. According to the equation:
d = ut + (1/2)at^2
where:
d = distance traveled
u = initial velocity
t = time taken
a = acceleration due to gravity
In this case, the palm fruit is falling vertically, so the initial velocity (u) is 0 m/s. The distance traveled (d) is 5m since it falls from a 5m tall tree. The acceleration due to gravity (a) is approximately 9.8 m/s^2.
Plugging these values into the equation, we have:
5 = 0 * t + (1/2) * 9.8 * t^2
Simplifying the equation, we get:
5 = 4.9t^2
Divide both sides of the equation by 4.9:
t^2 = 1
Taking the square root of both sides, we find:
t = 1 second
Therefore, it takes 1 second for the palm fruit to reach the ground.