Question

a rugby player passes the ball 5.34 m across the field, where it is caught at the same height as it left his hand. at what angle was the ball thrown if its initial speed was 7.7 m/s, assuming that the smaller of the two possible angles was used

Answer 1

`31.035^(\circ)`

Step-by-step explanation:

x = Displacement in x direction = 5.34 m

t = Time taken to travel the displacement

y = Displacement in y direction = 0

u = Initial velocity of ball = 7.7 m/s

g = Acceleration due to gravity =
`9.81\ \text{m/s}^2`

Displacement in x direction is given by

`x=u\cos\theta t\\\Rightarrow t=(5.34)/(7.7 \cos\theta)`

Displacement in y direction is given by

`y=u\sin\theta t-(1)/(2)gt^2\\\Rightarrow 0=7.7\sin\theta (5.34)/(7.7\cos\theta)-(1)/(2)* 9.81 ((5.34)/(7.7\cos\theta))^2\\\Rightarrow 0=7.7\sin\theta-4.905* (5.34)/(7.7\cos\theta)\\\Rightarrow 0=7.7^2\sin\theta \cos\theta-4.905* 5.34\\\Rightarrow 0=7.7^2(\sin2\theta)/(2)-4.905* 5.34\\\Rightarrow 0=7.7^2\sin2\theta-4.905*5.34* 2\\\Rightarrow \sin2\theta=(4.905* 5.34* 2)/(7.7^2)\\\Rightarrow 2\theta=\sin^(-1)(4.905* 5.34* 2)/(7.7^2)`

`\Rightarrow \theta=(62.07)/(2)\\\Rightarrow \theta=31.035^(\circ)`

The angle at which the ball was thrown is
`31.035^(\circ)`.