Respuesta :
Answer:
The time taken by the rock to reach the ground is 0.569 seconds.
Explanation:
Given that,
A student throws a rock horizontally off a 5.0 m tall building, s = 5 m
The initial speed of the rock, u = 6 m/s
We need to find the time taken by the rock to reach the ground. Using second equation of motion to find it. We get :
[tex]s=ut+\dfrac{1}{2}gt^2\\\\5=6t+\dfrac{1}{2}\times 9.8t^2\\\\t=0.569\ seconds[/tex]
So, the time taken by the rock to reach the ground is 0.569 seconds. Hence, this is the required solution.
Answer:
The rock to reach the ground in 0.569 sec.
Explanation:
Given that,
Height = 5.0 m
Speed = 6.0 m/s
A student throws a rock horizontally off a 0.5 m tall building.
This is the distance.
So, s = 0.5 m
We need to calculate the time
Using equation of motion
[tex]s=ut+\dfrac{1}{2}gt^2[/tex]
Where, s = distance
t = time
g = acceleration due to gravity
Put the value into the formula
[tex]5.0=6.0\times t+0.5\times9.8t^2[/tex]
[tex]4.9t^2+6.0 t-5.0 =0[/tex]
[tex]t = 0.569\ sec[/tex]
Neglect the negative value.
Hence, The rock to reach the ground in 0.569 sec.
