Question

a) An eagle is flying horizontally at 4.4 m/s with a fish in its claws. It accidentally drops the fish.

(i) How much time passes before the fish's speed triples?
___s

(ii) How much additional time would be required for the fish's speed to triple again?
___ s

b) A soccer player kicks the ball toward a goal that is 28.8 m in front of him. The ball leaves his foot at a speed of 19.7 m/s and an angle of 32.2° above the ground. Find the speed of the ball when the goalie catches it in front of the net. (Note: The answer is not 19.7 m/s.)
____m/s

c) A dolphin leaps out of the water at an angle of 36° above the horizontal. The horizontal component of the dolphin's velocity is 9.0 m/s. Find the magnitude of the vertical component of the velocity.
_____m/s

Answer 1

[A(i)] Given Velocity V_x=4.4m/s,

Acceleration g=9.8m/s²

Time t=? When velocity is v= 3×4.4=13.2m/s

Now since vertical velocity component will act , so we need to calculate it first.

So v_y=√[v²-v_x²]

V_y=√[(13.2)² –(4.4)²]

V_y=√[174.24–19.36]=12.44m/s

Now for time calculation we use

V_y=u+gt

12.44=0+9.8×t

t=12.44/9.8=1.26s

So fish's speed triple in 1.26s

(ii) now when velocity of fish triple again then it becomes

V=4.4×9=39.6m/s

Again we have to calculate vertical component of this velocity,which is

V_y=√[v²–v_x²]

V_y=√[(39.6)²–(4.4)²]

V_y=39.35m/s

Now for time

V_y=u+gt

39.35=0+9.8t

t=39.35/9.8=4.01s

So when fish's speed triple again time is 4.01s

(B) initial velocity by which ball leaves feet is u=19.7m/s

Horizontal distance is d=28.8m

Angle by which it hit is =32.2°

We have to calculate final Speed of the ball.

For that first we break the component of intial velocity in x and y direction

So u_x=19.7cos(32.2)=16.67m/s

Since ball is moving in projectile therefore final velocity in x is

v_x=Ux =16.67m/s

now in y direction u_y=19.7sin(32.2)=10.49m/s

time taken by ballvto cover the distance d is t=d/v_x

t=28.8/16.67=1.72s

now we need to calculate final velocity in y direction for which we use

V_y=u_y-gt

V_y=10.49-(9.8×1.72)=-6.36m/s

Now final speed by which golie catches it is

V=√[v_x²+v_y²]

V=√[(16.67)²+(-6.36)²]

V=√318.32=17.84m/s