Respuesta :
Answer:
[tex]t_t=4.131\ s[/tex]
Explanation:
Given:
height above the horizontal form where the ball is hit, [tex]y=1\ m[/tex]
angle of projectile above the horizontal, [tex]\theta=30^{\circ}[/tex]
initial speed of the projectile, [tex]u=40\ m.s^{-1}[/tex]
Firstly we find the vertical component of the initial velocity:
[tex]u_y=u.\sin\theta[/tex]
[tex]u_y=40\times \sin30^{\circ}[/tex]
[tex]u_y=20\ m.s^{-1}[/tex]
During the course of ascend in height of the ball when it reaches the maximum height then its vertical component of the velocity becomes zero.
So final vertical velocity during the course of ascend:
- [tex]v_y=0\ m.s^{-1}[/tex]
Using eq. of motion:
[tex]v_y^2=u_y^2-2g.h[/tex] (-ve sign means that the direction of velocity is opposite to the direction of acceleration)
[tex]0^2=20^2-2\times 9.8\times h[/tex]
[tex]h=20.4082\ m[/tex] (from the height where it is thrown)
Now we find the time taken to ascend to this height:
[tex]v_y=u_y-g.t[/tex]
[tex]0=20-9.8t[/tex]
[tex]t=2.041\ s[/tex]
Time taken to descent the total height:
- we've total height, [tex]h'=h+y[/tex] [tex]=20.4082+1[/tex]
[tex]h'=u_y'.t'+\frac{1}{2} g.t'^2[/tex]
- during the course of descend its initial vertical velocity is zero because it is at the top height, so [tex]u_y'=0\ m.s^{-1}[/tex]
[tex]21.4082=0+4.9t'^2[/tex]
[tex]t'=2.09\ s[/tex]
Now the total time taken by the ball to hit the ground:
[tex]t_t=t'+t[/tex]
[tex]t_t=2.09+2.041[/tex]
[tex]t_t=4.131\ s[/tex]
