Answer:
2.06 s
Explanation:
The boulder is in free fall, so it undergoes constant acceleration. A kinematic equation can be used to find the time it takes for the boulder to fall given the height of the cliff, the initial vertical velocity of the boulder, and the acceleration due to gravity.
The kinematic equation is:
s = ut + ½ at²
where s is the displacement,
u is the initial velocity,
a is the acceleration,
and t is time.
Given:
s = 20.7 m
u = 0 m/s
a = 9.8 m/s²
Find: t
20.7 = (0) t + ½ (9.8) t²
20.7 = 4.9 t²
t = 2.06
It takes 2.06 seconds for the boulder to reach the ground.
Answer:
2.06 seconds
Explanation:
To solve this problem, we can use the kinematic equation for vertical motion under constant acceleration:
[tex]\Large\boxed{\boxed{\sf d = v_0 t + \dfrac{1}{2} a t^2}} [/tex]
Where:
We want to find the time it takes for the boulder to reach the ground, so we need to solve for [tex]\sf t [/tex]. Rearranging the equation, we have:
[tex] \sf d = 0 \cdot t + \dfrac{1}{2} a t^2 [/tex]
[tex]\sf d = \dfrac{1}{2} a t^2 [/tex]
[tex]\sf 2d = a t^2 [/tex]
[tex]\sf t^2 = \dfrac{2d}{a} [/tex]
[tex]\sf t = \sqrt{\dfrac{2d}{a}} [/tex]
Now, let's substitute the given values into the equation:
[tex]\sf t = \sqrt{\dfrac{2 \times 20.7 \, \text{m}}{9.8 \, \text{m/s}^2}} [/tex]
[tex]\sf t = \sqrt{\dfrac{41.4 \, \text{m}}{9.8 \, \text{m/s}^2}} [/tex]
[tex]\sf t = \sqrt{4.2244897959183} [/tex]
[tex]\sf t \approx 2.0553563671340 [/tex]
[tex]\sf t \approx 2.06 \, \textsf{s (in 2 d.p.)} [/tex]
So, it takes approximately 2.06 seconds for the boulder to reach the ground.
