Question

A Stone is dropped from the top of a tower. When it has fallen a distance of 10m, another dropped from a point 38m below the top of the tower. If both the stones reach the ground at the same time, calculatea) The height of the tower andThe velocity of the stones when they reach the ground

Answer 1

Answer:

Step-by-step explanation:

Answer 2

Answer:

velocities of the stone 1 and 2 are 34.05 m/sec and 20 m/sec

Height of tower is 58m

Step-by-step explanation:

displacement of the first stone when second is dropped=5m,

thus time

difference in dropping of the stones is t=$\frac{2h}{g} =\frac{2*10}{10} =2sec\\\\\\ at u=0h= ut + 1/2gt^2 =0+1/2g(t-2)^2 +38\\\\ 1/2gt^2- 1/2gt^2=1/2(4g-4gt)+38\\\\t=2secs\\\\height of tower=38+1/2*10*2\\ \\ =58 metres$

mortion of the stone=

 $\sqrt{2gh} \\\\=\sqrt{2*10*10} \\\\=14m/sec$

velocity of the stone 1

v=$\sqrt{2*10*58} =34.05 m/sec$

velocity of stone 2

v=$\sqrt{2*10*(h-38)} \\\\=\sqrt{2*10*20}$$\sqrt{2*g*(h-38)} \\\\=\sqrt{2*10*(58-38)} \\\\=\sqrt{400} \\=20$

velocities of the stone 1 and 2 are 34.05 m/sec and 20 m/sec

Height of tower is 58m

#SPJ2