Answer:
(1) 14.12 m/s
Step-by-step explanation:
Given:
= initial speed of the ball = 16 m/s
= angle of the initial speed with the horizontal axis =
![28^\circ](https://img.qammunity.org/2020/formulas/physics/college/5kvramz9nx8b0nigowcsij16zlh20jw0sj.png)
= initial height of the ball from where Julie throws the ball = 1.5 m
= final position of the ball where Sarah catches the ball = 1.5 m
Let us assume the following:
= horizontal component of the initial speed
= vertical component of the initial speed
= horizontal acceleration of the ball
= vertical acceleration of the ball
The given problem is projectile motion. When the ball is thrown from the air with a speed of 16 m/s at an angle 28 degree with the horizontal axis. When the ball is in the air, it experiences an only gravitational force in the downward direction if we ignore air resistance on the ball.
This means if we break the motion of the ball along two axes and study it, we have a uniform acceleration motion in the vertical direction and a zero acceleration motion along the horizontal.
Since the ball has a zero acceleration motion along the horizontal axis, the ball must have a constant speed along the horizontal at all instant of time.
Let us find out the initial velocity horizontal component of the velocity of the ball. which is given by:
![u_x = u\cos 28^\circ = 16* \cos 28^\circ = 14.12\ m/s](https://img.qammunity.org/2020/formulas/physics/college/az5p7dp95820qsnw8rswmprelynhbq3gxo.png)
As this horizontal velocity remains constant in the horizontal motion at all instants of time. So, the horizontal component of the ball's velocity when Sarah catches the ball is 14.12 m/s.
Hence, the horizontal component of the ball's velocity when the ball is caught by Sarah is 14.12 m/s.