Question

A soccer ball is kicked from the ground at an angle of θ = 42 degrees with respect to the horizontal. The ball is in the air for a time tm = 1.3 s before it lands back on the ground.Numerically, what is the total horizontal distance, dm in meters, traveled by the ball in the time, tm?

Answer 1

Answer:

Explanation:

Given

Launch angle $\theta=42^{\circ}$

time period of flight $T=1.3\ s$

Time period of Projectile $T=\frac{2u\sin \theta }{g}$

Where u=initial velocity

g=acceleration due to gravity

$1.3=\frac{2\times u\times \sin (42)}{9.8}$

$u=9.51\ m/s$

Range of Projectile is given by

$R=\frac{u^2\sin 2\theta }{g}$

$R=\frac{9.51^2\times \sin (84)}{9.8}$

$R=\frac{90.44\times 0.9945}{9.8}=9.17\ m$

Answer 2

The total horizontal distance covered by ball is of 9.17 m.

Given data:

The angle with respect to the horizontal is, θ = 42 degrees.

The time of flight of ball is, t = 1.3 s.

The concept of projectile motion is applied to solve this problem. The motion having some inclination is known as projectile motion. So, the time of flight of ball is,

$t=\dfrac{2usin\theta}{g}$

here, u is the initial velocity of ball. Solving as,

$1.3=\dfrac{2 \times u \times sin42}{9.8}\\\\u = 9.51 \;\rm m/s$

Now, the the total horizontal distance covered by ball is called Range. So, the expression for the range is,

$R=\dfrac{u^{2}sin2 \theta}{g}\\\\R=\dfrac{9.51^{2} \times sin2 (42)}{9.8}\\\\R= 9.17 \;\rm m$

Thus, we can conclude that the total horizontal distance covered by ball is of 9.17 m.

Learn more about the projectile motion here:

https://brainly.com/question/11049671