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A ball is released from a height and it reaches the ground in 3s. If g=9.8m/s^2 , find:

(a) The height from which the ball was released,

(b) The velocity with which the ball will strike the ground.

Answer:

Given: t=3s, g = 9.8m/s^2 , initial velocity u=0m/s

Assume ‘s’ to be the height the ball is released;

(a)We know from the equation of motion;

\begin{aligned} S &=u t+1 / 2 a t^{2} \\ &=0+1 / 2(9.8)(3)^{2} \\ &=44.1 \mathrm{m} \end{aligned}

(b) Assume ‘v’ to be the velocity with which the ball strikes the ground,

We know from the equation of motion,

\begin{array}{l} v^{2}-u^{2}=2 g s \\ v^{2}-0^{2}=2(9.8)(44.1) \\ v^{2}=864.36 \\ v=29.4 \mathrm{m} / \mathrm{s} \end{array}

hello everyone welcome to lido learning i'm good preet your science tutor today's question is a ball is released from height and it reaches the ground in three seconds if g is equals to 9.8 per meter per second square fine a the height from which the ball was released and b the velocity with which the ball will strike the ground so let's start our answer here the given is time t t is equals to 3 seconds okay and g is given as 9.8 meter per second square initial velocity is zero that is u is equals to zero so for the calculation of height we will use the equation of motion that is s is equals to u d plus half a t square so from here we can calculate the value of s or you can say the height the first term will be cancelled as u is zero so you will get as half acceleration is nine point a and t is 3 square so you will get the value as 44.1 meter as the value of height now for the second part that is b part you have to calculate the velocity so velocity can be calculated by using the third equation of motion v square minus u square is equals to 2 g s okay so here v square you have to calculate and u is already 0 so you have to put the values in 2gs 2 into 9.8 into s the value of s is 44.1 which we have already calculated in the part a so from here v square is equals to 864.36 and v is equals to 29.4 meter per second so these are your answers for more such videos please subscribe leader learning and for any doubt drop a comment thank you

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