Question

A second basemen tosses the ball to the first basemen, who catches it at the same level from which it was thrown. The throw is made with an initial speed of 19.0 m/s at an angle of 34.5 degrees above the horizontal.

what is the horizontal component of the balls velocity just before it is caught?
how long is the ball in the air?

Answer 1

Answer:

Explanation:

Given

Initial speed of ball $u=19\ m/s$

Launch angle $\theta =34.5^{\circ}$

As there is no acceleration in horizontal direction therefore there is no change in velocity

thus horizontal velocity remains unchanged

$u_x=u\cos \theta$

$u_x=19\cdot \cos 34.5$

$u_x=15.65\ m/s$

Time of flight of projectile is

$T=\frac{2u\sin \theta }{g}$

$T=\frac{2\times 19\times \sin (34.5)}{9.8}$

$T=2.196\ s$

Thus time for which it is in air is 2.12 s                        

Answer 2

Answer:

Explanation:

velocity of projection, u = 19 m/s

angle of projection, θ = 34.5°

The horizontal component of velocity remains same as there is no acceleration in the horizontal direction.

So, horizontal component of velocity is given by

ux = u Cos θ

ux = 19 Cos 34.5

ux = 15.66 m/s

Let the ball remains in air for time T.

Use the formula for time of flight.

$T = \frac{2uSin\theta }{g}$

$T = \frac{2\times 19Sin34.5}{9.8}$

t = 2.2 seconds