Question

From the top of a tall building (height 27.3 m), a boy throws an apple upward, which strikes ground after 16 s. Take g= 9.8 m/s², find the speed of apple with which it was thrown and the maximum height reached by it.​

Answer 1

Let the top of the building be taken as origin of coordinate system. The upward direction be taken as positive and downward direction as negative. The ball returns to ground after t= 16 s, say with a velocity u m/s upwards, and is acted by acceleration due to gravity= -9.8 m/s². We can find the magnitude of u using the relation:

s = ut +½ a t².

Here s= -27.3 m, a= 9.8 m/s², and t = 16 s

Substituting these values we get,

$- 27.3 = 16u - \frac{1}{2} \times 9.8 \times {16}^{2} \\ - 27.3 = 16u - 1254.4 \\ 16u = 1254.4 - 27.3 \\ u = \frac{1227.1}{16} = 76.7m {s}^{ - 1}$

Now,

${v}^{2} = {u}^{2} + 2as \\ 0 = {76.7}^{2} + 2( - 9.8)s \\ s = \frac{5882.89}{19.6} = 300.14m$

Hope It Helps :)