Answer:
[tex]38.3\ \text{m/s}[/tex]
[tex]32.14\ \text{m/s}[/tex]
6.55 seconds
[tex]52.65\ \text{m}[/tex]
[tex]254.84\ \text{m}[/tex]
Explanation:
u = Initial velocity of rock = 50 m/s
[tex]\theta[/tex] = Angle of throw = [tex]40^{\circ}[/tex]
g = Acceleration due to gravity = [tex]9.81\ \text{m/s}^2[/tex]
Horizontal component is given by
[tex]u_x=u\cos\theta\\\Rightarrow u_x=50\times \cos40^{\circ}\\\Rightarrow u_x=38.3\ \text{m/s}[/tex]
The horizontal component of the velocity is [tex]38.3\ \text{m/s}[/tex]
Vertical component is given by
[tex]u_y=u\sin\theta\\\Rightarrow u_y=50\times \sin40^{\circ}\\\Rightarrow u_y=32.14\ \text{m/s}[/tex]
The horizontal component of the velocity is [tex]32.14\ \text{m/s}[/tex]
Time of flight is given by
[tex]t=\dfrac{2u\sin\theta}{g}\\\Rightarrow t=\dfrac{2\times 50\sin40^{\circ}}{9.81}\\\Rightarrow t=6.55\ \text{s}[/tex]
The hang time of the rock is 6.55 seconds
Maximum height is given by
[tex]h=\dfrac{u^2\sin^2\theta}{2g}\\\Rightarrow h=\dfrac{50^2\sin^240^{\circ}}{2\times 9.81}\\\Rightarrow h=52.65\ \text{m}[/tex]
Maximum height is [tex]52.65\ \text{m}[/tex]
Range is given by
[tex]d=\dfrac{u^2\sin2\theta}{g}\\\Rightarrow d=\dfrac{50^2\sin(2\times40)^{\circ}}{9.81}\\\Rightarrow d=254.84\ \text{m}[/tex]
The range is [tex]254.84\ \text{m}[/tex]
