Answer :
Answer:
94.13 ft/s
Explanation:
Given:
- [tex]t[/tex] = time interval in which the rock hits the opponent = 10 s - 5 s = 5 s
- [tex]s[/tex] = distance to be moved by the rock long the horizontal = 98 yards
- [tex]y[/tex] = displacement to be moved by the rock during the time of flight along the vertical = 0 yard
Assume:
- [tex]u[/tex] = magnitude of initial velocity of the rock
- [tex]\theta[/tex] = angle of the initial velocity with the horizontal.
For the motion of the rock along the vertical during the time of flight, the rock has a constant acceleration in the vertically downward direction.
[tex]\therefore y = u\sin \theta t +\dfrac{1}{2}(-g)t^2\\\Rightarrow 0 = u\sin \theta 5 +\dfrac{1}{2}(-9.8)\times 5^2\\\Rightarrow u\sin \theta 5 =\dfrac{1}{2}(9.8)\times 5^2......(1)\\[/tex]
Now the rock has zero acceleration along the horizontal. This means it has a constant velocity along the horizontal during the time of flight.
[tex]\therefore u\cos \theta t = s\\\Rightarrow u\cos \theta 5 = 98.....(2)\\[/tex]
On dividing equation (1) by (2), we have
[tex]\tan \theta = \dfrac{25}{20}\\\Rightarrow \tan \theta = 1.25\\\Rightarrow \theta = \tan^{-1}1.25\\\Rightarrow \theta = 51.34^\circ[/tex]
Now, putting this value in equation (2), we have
[tex]u\cos 51.34^\circ\times 5 = 98\\\Rightarrow u = \dfrac{98}{5\cos 51.34^\circ}\\\Rightarrow u =31.38\ yard/s\\\Rightarrow u =31.38\times 3\ ft/s\\\Rightarrow u =94.13\ ft/s[/tex]
Hence, the initial velocity of the rock must a magnitude of 94.13 ft/s to hit the opponent exactly at 98 yards.