Question

A rugby player passes the ball 7.75 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 12.0 m/s, assuming that the smaller of the two possible angles was used? Number What other angle gives the same range? Number For the pass using the smaller of the two angles, how long is the ball in the air?

Answer 1

15.93°

74.06°

0.67 seconds

Step-by-step explanation:

Angle at which the projectile is shot at = θ

g = Acceleration due to gravity = 9.81 m/s²

Range of projectile

`R=\frac {v^(2)\sin 2\theta}{g}\\\Rightarrow \theta=(1)/(2)\sin^(-1)\left((Rg)/(v^2)\right)\\\Rightarrow \theta=(1)/(2)\sin^(-1)\left((7.75* 9.81)/(12^2)\right)\\\Rightarrow \theta=15.93^(\circ)`

Hence, the angle at which the ball was thrown is 15.93° or 90-15.93 = 74.06°

Angle at which the projectile is shot at = θ = 15.93°

`t=(2v\sin(\theta))/(g)\\\Rightarrow t=(2* 12\sin(15.93))/(9.81)\\\Rightarrow t=0.67\ s`

Time the ball was in the air is 0.67 seconds