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

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

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

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