During soccer practice, Maya kicked a soccer ball 37° off the ground at 25 m/s. Max, the goalie, caught the ball 60 m away from

Question

Masja [62]

3 years ago

9

During soccer practice, Maya kicked a soccer ball 37° off the ground at 25 m/s. Max, the goalie, caught the ball 60 m away from

Maya and 1.0 m off the ground. Max caught the ball ______ s after Maya kicked it.

Physics

2 answers:

amid [387] · Answer 1 · 2022-02-23T23:58:30+03:00

Answer:

3.0 seconds

Explanation:

We can solve the problem by considering the horizontal motion of the ball only. In fact, the ball moves by uniform motion (constant speed) along the horizontal direction, since there are no forces acting in this direction. The horizontal speed of the ball is given by:

$v_x = v_0 cos \theta = (25 m/s)(cos 37^{\circ})=19.97 m/s$

and it does not change during the motion.

We also know that the ball travels a horizontal distance of d = 60 m, so we can find the time it takes to cover the distance by using the equation:

$t=\frac{d}{v}=\frac{60 m}{19.97 s}=3.0 s$

n200080 [17] · Answer 2 · 2022-02-24T02:10:08+03:00

Max will catch the ball $\fbox{3\,{\text{s}}}$ after it was kicked by Maya.

Further Explanation:

The soccer ball kicked by Maya at a particular angle from the horizontal moves in the two-dimensional plane. It covers the distance in the vertical as well as in the horizontal direction.

As this motion of the soccer ball is a projectile motion, the vertical direction of motion of the ball is the motion under the action of acceleration due to gravity whereas the motion of the ball in the horizontal direction is the linear motion of the ball.

The velocity of the ball in the horizontal direction:

$\fbox{\begin\\{v_x}=v\cos\,37^\circ\end{minispace}}$

Here, ${v_x}$ is the speed of the ball in the horizontal direction.

Substitute the value of $v$ in above expression.

$\begin{aligned}{v_x}&=\left({25\,{{\text{m}}\mathord{\left/{\vphantom{{\text{m}}{\text{s}}}}\right.\kern-\nulldelimiterspace}{\text{s}}}}\right)\left({\frac{4}{5}}\right)\\&=20\,{{\text{m}}\mathord{\left/{\vphantom{{\text{m}}{\text{s}}}}\right.\kern-\nulldelimiterspace}{\text{s}}}\\\end{aligned}$

The motion of the ball in the horizontal direction is a linear motion and during this, the speed of the ball in the horizontal direction remains the same.

Since, the ball covers $60\,{\text{m}}$ in horizontal direction from Maya to reach to Max at the goalkeeping end, the ball will travel this distance at the speed of $20\,{{\text{m}} \mathord{\left/ {\vphantom {{\text{m}} {\text{s}}}} \right. \kern-\nulldelimiterspace} {\text{s}}}$ .

The time taken by the ball to reach Max at the goalkeeping end is:

$\begin{aligned}t&=\frac{{{\text{distance}}}}{{{\text{speed}}}}\\&=\frac{{60\,{\text{m}}}}{{20\,{{\text{m}}\mathord{\left/{\vphantom{{\text{m}}{\text{s}}}}\right.\kern-\nulldelimiterspace}{\text{s}}}}}\\&=3\,{\text{s}}\\\end{aligned}$