Question

Barack is playing basketball in his back yard. He takes a shot 7.0 m from the basket (measured along the ground), shooting at an angle of 45 degrees. The ball is 2.0 m off the ground when it leaves his hand, and hits the backboard 3.5 m off the ground. Neglecting air friction, about how long is the ball in flight?

Answer 1

The time of flight of the ball is 1.06 seconds.

Step-by-step explanation:

Given
`\Delta x=7\ m`

`\theta=45 \°`

Also,
`\Delta y=(3.5-2)=1.5\ m`

`a_x=0\ and\ a_y=-9.81\ m/s^2`

Let us say the velocity in the x-direction is
`v_x` and in the y-direction is
`v_y`. And acceleration in the x-direction is
`a_x` and in the y-direction is
`a_y`.

Also,
`\Delta x\ and\ \Delta y` is distance covered in x and y direction respectively. And
`t` is the time taken by the ball to hit the backboard.

We can write
`v_x=v_0cos(45)\ and\ v_y=v_0sin(45)`. Where
`v_0` is velocity of ball.

Now,

`\Delta x=v_x* t+(1)/(2)* a_x* t^2\\ \Delta x=v_x* t+(1)/(2)* 0* t^2\\\Delta x=v_xt`

`\Delta x=v_0cos(45)* t\\7=v_0cos(45)* t\\\\t=(7)/(v_0cos(45))`

Also,

`\Delta y=v_y* t+(1)/(2)* a_y* t^2\\ 1.5=v_0sin(45)* (7)/(v_0cos(45))+(1)/(2)* (-9.81)*((7)/(v_0cos(45)) )^2\\\\1.5=7-(481)/((v_0)^2)\\ \\(481)/((v_0)^2)=5.5\\\\(v_0)^2=(481)/(5.5)\\ \\(v_0)^2=87.45\\\\v_0=√(87.45)=9.35\ m/s`.

Plugging this value in

`t=(7)/(v_0cos(45))\\ \\t=(7)/(9.35* 0.707)\\ \\t=(7)/(6.611)`

`t=1.06\ seconds`

So, the time of flight of the ball is 1.06 seconds.