Answer:
[tex]28.53\ \text{m}[/tex]
[tex]4.48\ \text{s}[/tex]
[tex]72.1\ \text{m}[/tex]
Explanation:
[tex]y_0[/tex] = Height from where the ball is thrown = [tex]70\ \text{m}[/tex]
[tex]\theta[/tex] = Angle of inclination = [tex]45^{\circ}[/tex]
[tex]u[/tex] = Initial velocity = [tex]9\ \text{m/s}[/tex]
[tex]g[/tex] = Acceleration due to gravity = [tex]9.8\ \text{m/s}^2[/tex]
Range of projectile is given by
[tex]R=\dfrac{u\cos \theta }{g}\left(u\sin \theta +{\sqrt{u^{2}\sin ^{2}\theta +2gy_{0}}}\right)\\\Rightarrow R=\dfrac{9\cos45^{\circ}}{9.8}\left(9\sin45^{\circ}+\sqrt{9^2\left(\sin45^{\circ}\right)^2+2\times 9.8\times 70}\right)\\\Rightarrow R=28.53\ \text{m}[/tex]
The ball will land [tex]28.53\ \text{m}[/tex] away.
Time of flight is given by
[tex]t=\dfrac{u\sin \theta +{\sqrt{(u\sin \theta )^{2}+2gy_{0}}}}{g}\\\Rightarrow t=\dfrac{9\sin 45^{\circ} +{\sqrt{(9\sin 45^{\circ} )^{2}+2\times 9.8\times 70}}}{9.8}\\\Rightarrow t=4.48\ \text{s}[/tex]
The time the ball will stay in the air is [tex]4.48\ \text{s}[/tex]
Maximum height is given by
[tex]h=y_0+\dfrac{u^2\sin^2\theta}{2g}\\\Rightarrow h=70+\dfrac{9^2\times \sin^245^{\circ}}{2\times 9.8}\\\Rightarrow h=72.1\ \text{m}[/tex]
Maximum height the object will reach above the ground is [tex]72.1\ \text{m}[/tex].
