Question

A ball is thrown directly downward with an initial speed of 8.45 m/s, from a height of 29.9 m. After what time interval does it strike the ground?

Answer 1

Solution:

Let us go to the basics first.

From Newton’s Equation of motion, we know that:

S = ut + (1/2)at²

[S = height = 29.9 m; u = initial speed = 8.45 m/s; a=acceleration = g = 9.81 m/s²]

=> 29.9 = 8.45 t + (1/2) 9.81t²

=> 29.9 = 8.45 t + 4.905t²

=> 4.905t² + 8.45 t - 29.9 = 0

Solving the above quadratic equation,

t= -3.4763   or   1.7535

since time cannot be negative, so t = 1.7535 seconds         (Answer)

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