Answer to Question #204486 in Physics for Jasmin

Question

A football player kicks a field goal, launching the ball at an angle of 49∘ above the horizontal. During the kick, the ball is in contact with the player's foot for 0.049 s, and the ball's acceleration is 260 m/s^2. What is the range of the football?

Express your answer to two significant figures and include appropriate units.

Answer 1

Under the acceleration a = 260 m/s^2 in time t = 0.049s the ball reaches the following ininial speed:

$$v_0 = at = 260\cdot 0.049 = 12.74m/s$$

The range is given by the following formula (see https://courses.lumenlearning.com/boundless-physics/chapter/projectile-motion/):

$$R = \dfrac{v_0^2\sin 2\theta}{g}$$

where $\theta = 49\degree$ is the launching angle, and $g = 9.8m/s^2$ is the gravitational acceleration. Thus, obtain:

$$R = \dfrac{12.74^2\cdot \sin (2\cdot 49\degree)}{9.8} \approx 16.4m$$

Answer. 16.4 m.