Respuesta :
Answer:
[tex]u'=10.259\ m.s^{-1}[/tex] is the initial velocity of tossing the apple.
the apple should be tossed after [tex]\Delta t=0.0173\ s[/tex]
Explanation:
Given:
- velocity of arrow in projectile, [tex]v=30\ m.s^{-1}[/tex]
- angle of projectile from the horizontal, [tex]\theta=20^{\circ}[/tex]
- distance of the point of tossing up of an apple, [tex]d=30\ m[/tex]
Now the horizontal component of velocity:
[tex]v_x=v\ cos\ \theta[/tex]
[tex]v_x=30\times cos\ 20^{\circ}[/tex]
[tex]v_x=28.191\ m.s^{-1}[/tex]
The vertical component of the velocity:
[tex]v_y=v.sin\ \theta[/tex]
[tex]v_y=30\times sin\ 20^{\circ}[/tex]
[tex]v_y=10.261\ m.s^{-1}[/tex]
Time taken by the projectile to travel the distance of 30 m:
[tex]t=\frac{d}{v_x}[/tex]
[tex]t=\frac{30}{28.191}[/tex]
[tex]t=1.0642\ s[/tex]
Vertical position of the projectile at this time:
[tex]h=v_y.t-\frac{1}{2}g.t^2[/tex]
[tex]h=10.261\times 1.0642-\frac{1}{2} \times 9.8\times 1.0642^2[/tex]
[tex]h=5.3701\ m[/tex]
Now this height should be the maximum height of the tossed apple where its velocity becomes zero.
[tex]v'^2=u'^2-2g.h[/tex]
[tex]0^2=u'^2-2\times 9.8\times 5.3701[/tex]
[tex]u'=10.259\ m.s^{-1}[/tex] is the initial velocity of tossing the apple.
Time taken to reach this height:
[tex]v'=u'-g.t'[/tex]
[tex]0=10.259-9.8\times t'[/tex]
[tex]t'=1.0469\ s[/tex]
We observe that [tex]t>t'[/tex] hence the time after the launch of the projectile after which the apple should be tossed is:
[tex]\Delta t=t-t'[/tex]
[tex]\Delta t=1.0642-1.0469[/tex]
[tex]\Delta t=0.0173\ s[/tex]
