Question

A ball is thrown horizontally at a height of 1.6 metres above the ground, with initialspeed 14 m s^-1a)Find the time of flight of the ball, giving the answer as a fraction.b)Find the range of the ball.

Answer 1

Given:

The initial height of the ball, h=1.6 m

The initial speed of the ball, u=14 m/s

To find:

a) The time of flight of the ball.

b) The range of the ball.

Step-by-step explanation:

As the ball is thrown horizontally, the ball will have no vertical component of the initial velocity. The velocity of the ball is completely horizontal.

Thus the vertical component of the initial velocity of the ball is u_y=0 m/s.

The horizontal component of the initial velocity of the ball is u_x=u=14 m/s.

a)

From the equation of motion,

`h=u_yt+(1)/(2)gt^2`

Where g is the acceleration due to gravity and t is the time of flight of the ball.

On substituting the known values,

`\begin{gathered} 1.6=0+(1)/(2)*9.8* t^2 \\ \implies t=\sqrt{(2*1.6)/(9.8)} \\ =0.57\text{ s} \end{gathered}`

b)

The range of the ball is given by,

`R=u_xt`

On substituting the known values,

`\begin{gathered} R=14*0.57 \\ =7.98\text{ m} \end{gathered}`

Final answer:

a) The time of flight of the ball is 0.57 s

b) The range of the ball is 7.98 m