Question

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 of 4 m/s and the end of the slide is 0.7 m above the ground, her final speed will be: (in m/s)

Answer 1

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.

$v^{2} - u^{2} = 2aS$

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,

$v^{2} = u^{2} + 2aS$

$v^{2} = 4^{2} + 2 \times 9.8 \times 1.4$

$v^{2}$ = 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