Question

A body falls freely from rest.Find(a)its acceleration(b)the distance it falld in 3s.(c)its speed after falling 70m.(d)the time required to reach a speed of 25ms .(e)the time taken to fall 300m.

Answer 1

Answer:

a = 9.8 , s = 44.1 m , V = 37 m/s , t = 2.5 s , t = 7.75 s

( answers to respective questions )

Explanation:

earth attracts every body with the acceleration of 9.8 m/s2 .

acceleration will be 9.8 m/s2 .

Answer 2

Given:

The body starts to fall freely from rest.

To Find:

(a) Its acceleration.

(b) The distance it fell in 3 seconds.

(c) Its speed after falling 70 m.

(d) The time required to reach a speed of 25 ms⁻¹.

(e) The time taken to fall 300 m.

Solution:

→ Assume: Acceleration due to gravity (g) = 10 ms⁻²

→ The body falls from rest, therefore initial velocity (u) is equal to zero.

∴ Initial velocity (u) = 0

→ We will assume the downwards direction of acceleration and velocity to be positive.

(a) The acceleration of body (a) = g = 10 ms⁻²

(b) Distance fell in 3 seconds.

    Time (t) = 3 sec        

    Initial velocity (u) = 0

    Acceleration (a) = 10 ms⁻²

   $By\:using\:the\;2nd\:Equation\:of\:motion:\\\\S=ut+\frac{1}{2}at^{2} \\\\S=0+\frac{1}{2}(g)t^{2} \\\\S=\frac{1}{2}\times10\times(3)^{2} =45\:m$

∴ The distance fell in 3 seconds is equal to 45 m.

(c) Its speed after falling 70 m.

    Distance traveled (S) = 70 m

    Initial velocity (u) = 0

    Acceleration (a) = 10 ms⁻²

    $By \:using\:the\:3rd\:Equation\:of\:motion:\\\\v^{2} =u^{2}+2aS\\\\ v^{2} =0^{2}+2\times g \times S\\\\v^{2}=2 \times(10)\times70\\\\v^{2} =1400\\\\v=\sqrt{1400} =37.42\:ms^{-1}$

∴ Its speed after falling 70 m is equal to 37.42 ms⁻¹.

(d) Time required to reach a speed of 25 ms⁻¹

    Final velocity (v) = 25 ms⁻¹

    Initial velocity (u) = 0

    Acceleration (a) = 10 ms⁻²

    $By \:using\:the\:1st\:Equation\:of\:motion:\\\\v=u+at\\\\25 = 0+gt\\\\25=10\times t\\\\t=2.5\:seconds$

∴ The time required to reach a speed of 25 ms⁻¹ is equal to 2.5 seconds.

(e) The time taken to fall 300 m.

    Distance traveled (S) = 300 m

    Initial velocity (u) = 0

    Acceleration (a) = 10 ms⁻²

   $By\:using \:the\:2nd\:Equation\:of\:Motion:\\\\S=ut+\frac{1}{2}at^{2} \\\\300=(0)+\frac{1}{2}\times10\times t^{2} \\\\300=5t^{2}$

  $\therefore t^{2}=60\\\\ \therefore t=\sqrt{60}=7.75\:seconds$

∴ The time taken to fall 300 m is equal to 7.75 seconds.

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