Question

A basketball player is standing on the floor 100 m from the basket as in the figure below. The height of the basket is H - 3.05 m, and he shoots the ball at an angle, 0-38.0°, with the horizontal from a height of h = 1.84 m. (a) What is the acceleration of the basketball at the highest point in its trajectory? magnitude m/s direction -Solo... (b) At what speed must the player throw the basketbal so that the ball goes through the hoop without striking the backboard? m/s

Answer 1

( a )

Throughout it's trajectory the basket ball experiences an uniform acceleration   due to the force of gravity $=m.g$ acting downwards,

So, since the constant gravitational force acts on the ball at every point in its trajectory the acceleration of the ball due to force of gravity acting at the highest point in it's trajectory is,

magnitude =

g=9.8 m/s

direction = downward

( b )

equation for the trajectory of basket ball is given as,

$y=x.tan(\theta)-\frac{g.x^{2}}{2\:v_{0}^{2}\:cos^{2}(\theta)}$ _____________________________________relation 1

where,

$x=$ horizontal distance the ball has traveled measured with respect to the point from where it is thrown.

$y=$ vertical distance of the ball when it has traveled horizontal distance , measured with respect with respect to horizontal axis passing through the point of release.

$\theta=$ the angle with respect to horizontal with which the ball has been shot

$v_{0}=$ speed of the ball with which it has to be thrown

$g=$ acceleration due to gravity ( acceleration of the ball over its entire trajectory due to the gravitation force acting

on it downwards.)

assuming the ball goes into basket we have,

horizontal distance traveled by the bal = separation between basket and player

that is, $x=10\:m$

vertical distance of the ball at the horizontal position of $x=10\:m$ measured as mentioned earlier is given as,

$y=$ height of the basket  $(H=3.05\:m)$  height of the person $(h=1.84\:m)$

$y=H-h=3.05-1.84=1.21\:m$

$\theta=38.0\degree$

g=9.8 m/s

Using these values in relation 1 we get,

$y=x.tan(\theta)-\frac{g.x^{2}}{2\:v_{0}^{2}\:cos^{2}(\theta)}=10.tan(38\degree)-\frac{9.8\times10^{2}}{2\:v_{0}^{2}\:cos^{2}(38\degree)}$

$1.21=7.8129-\frac{789.099}{2\:v_{0}^{2}}$

$\frac{789.099}{2\:v_{0}^{2}}=7.8129-1.21$

$\frac{789.099}{2\:v_{0}^{2}}=6.6029$

$\frac{789.099}{2\times6.6029}=v_{0}^{2}$

$\sqrt{\frac{789.099}{2\times6.6029}}=v_{0}$

$\sqrt{\frac{789.099}{13.2058}}=v_{0}$

$\sqrt{59.75}=v_{0}$

$7.73=v_{0}$

Therefore the player must throw the ball with initial speed of