A child slides down a frictionless playground slide from a height of 2.1 m above the ground. If she starts with an initial speed
1 answer:
Answer: Final speed will be 6.59 m/s
Explanation: Initial height of slide is 2.1 m
Final height = 0.7 m
Initial speed, u = 4 m/s
Final speed, v = ??
To find the final speed in this question we can use equation of motion i.e.
Here, S is the distance covered by the child on the slide i.e. net height
S = 2.1 - 0.7 = 1.4 m
and a = g, acceleration due to gravity i.e. 9.8 m/s
Put all the values in the equation,
= 16 + 27.44 = 43.44
Taking square root both the sides, we get
v = 6.59 m/s
Final speed will be 6.59 m/s
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Answer:
d = 0.38 m
Explanation:
As we know that the person due to the airbag action, comes to a complete stop, in 36 msec or less, and during this time, is decelerated at a constant rate of 60 g, we can find the initial velocity (when airbag starts to work), as follows:
vf = v₀ -a*t
If vf = 0, we can solve for v₀:
v₀ = a*t = 60*9.8 m/s²*36*10⁻³s = 21.2 m/s
With the values of v₀, a and t, we can find Δx, applying any kinematic equation that relates all of some of these parameters with the displacement.
Just for simplicity, we can use the following equation:
where vf=0, v₀ =21.2 m/s and a= -588 m/s².
Solving for d:
⇒ d = 0.38 m