Respuesta :
Answer:
a. [tex]t=2.997\ s[/tex] is the time after which the two stone hit the water surface.
b. [tex]u_2=1.998\ m.s^{-1}[/tex]
c. [tex]v_1=31.3688\ m.s^{-1}[/tex] & [tex]v_2=31.3686\ m.s^{-1}[/tex]
Explanation:
Given:
- height of the cliff, [tex]h=50\ m[/tex]
- time gap between the projection of stones, [tex]\Delta t=1\ s[/tex]
- initial velocity of first stone, [tex]u_1=+2\ m.s^{-1}[/tex]
Here positive sign means that the stone is thrown vertically downward in the direction of gravity.
a.
Using equation of motion:
[tex]h=u_1.t+\frac{1}{2} g.t^2[/tex]
[tex]50=2t+4.9t^2[/tex]
[tex]t=2.997\ s[/tex] is the time after which the two stone hit the water surface.
b.
using the equation of motion for second stone:
[tex]h=u_2.t+\frac{1}{2} g.t^2[/tex]
[tex]50=u_2\times 2.997+4.9\times 2.997^2[/tex]
[tex]u_2=1.998\ m.s^{-1}[/tex]
c.
Final velocity of the stone at the instant of touching the water surface:
[tex]v_1^2=u_1^2+2.g.h[/tex]
[tex]v_1^2=2^2+2\times 9.8\times50[/tex]
[tex]v_1=31.3688\ m.s^{-1}[/tex]
&
[tex]v_2^2=u_2^2+2g.h[/tex]
[tex]v_2^2=1.998^2+2\times 9.8\times 50[/tex]
[tex]v_2=31.3686\ m.s^{-1}[/tex]
