Name: Speed and Velocity in Daily Life
Uploaded: 2019-07-19
Duration: 2 min 18 s
Description: Our life would be nothing without motion! So, let's dig a little deeper and see what motion entails. Watch this video to learn all about speed and velocity.

Question

A golfer hits a golf ball into the air over level ground. The velocity of the ball at a height of $10.3 m$ is $\overset{⇀}{v} = (8.6 \hat{i} + 7.2 \hat{j}) m s^{- 1}$ , with $\hat{i}$ horizontal and $\hat{j}$ upward. Find

$(a)$ the maximum height of the ball and

$(b)$ the total horizontal distance traveled by the ball, what is the

$(c)$ magnitude and

$(d)$ angle (below the horizontal) of the ball's velocity just before it touches the ground?

Accepted Answer

A golfer hits a golf ball into the air over level ground at an initial speed of $v_{0}$ and angle $θ_{0}$ .

The velocity of the ball at a height of $10.3 m$ is $\overset{⇀}{v} = (8.6 \hat{i} + 7.2 \hat{j}) m s^{- 1}$ , with $\hat{i}$ horizontal and $(b)$ upward.

The horizontal component of the velocity throughout the trajectory remains constant at,

$v_{x} = v_{0} \cos (θ_{0}) . . . (1)$

The vertical component of the velocity and the height are related as,

$2 y g = {\{v_{0} \sin (θ_{0})\}}^{2} - {v_{y}}^{2} . . . (2)$

$(a)$ Substituting the given values into equations $(1)$ and $(2)$

$v_{x} = v_{0} \cos (θ_{0}) = 8.6 . . . (3)$

$2 \times 10.3 \times 9.8 = {\{v_{0} \sin (θ_{0})\}}^{2} - 7 . 2^{2}$

$\Rightarrow {\{v_{0} \sin (θ_{0})\}}^{2} = 253.7$

$\Rightarrow v_{0} \sin (θ_{0}) = 15.9 . . . (4)$

Dividing equation $(4)$ with $(5)$

$\tan (θ_{0}) = \frac{15.9}{8.6} = 1.85$

$\Rightarrow θ_{0} = \tan^{- 1} (1.85) = 61.5 °$

The maximum height of the ball,

$h_{\max} = \frac{{\{v_{0} \sin (θ_{0})\}}^{2}}{2 g}$

$= \frac{253.7}{2 \times 9.8} = 12.9 m$

$(b)$ Total horizontal distance travelled by the ball,

$R = \frac{2 \{v_{0} \cos (θ_{0})\} \{v_{0} \sin (θ_{0})\}}{g}$

$= \frac{8.6 \times 15.9}{9.8} = 27.8 m$

$(c)$ The magnitude of the ball's velocity just before it touches the ground,

$v_{0} = \sqrt{{\{v_{0} \cos (θ_{0})\}}^{2} + {\{v_{0} \sin (θ_{0})\}}^{2}}$

$= \sqrt{8 . 6^{2} + 15 . 9^{2}} = 18.1 m s^{- 1}$

$(d)$ As calculated in equation $(5)$ the angle would be, $61.5 °$ .

Important Questions on Motion in Two and Three Dimensions

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