A pickup truck moves at 25 m/s toward the east. Ahmed is standing in the back and throws a baseball in what to him is the southwest direction at 28 m/s (with respect to the truck). A person at rest on the ground would see the ball moving how fast in what direction? HTML EditorKeyboard Shortcuts

Speed = 20 m/sec at 75 deg South of East = 20 m/sec at 15 deg East of South

Explanation:

given data

truck moves = 25 m/s toward the east.

throws a baseball = 28 m/s southwest

solution

first we take here Speed of truck w.r.to ground i.e. V(p/g) = 25 m/sec toward the east  so we can say

V(p/g) = (25 i) m/sec     ........................1

and

Speed of baseball w.r.t. pickup i.e. V(b/p) = 28 m/sec toward the South West  and   we know that south west direction is in third quadrant

and here both component (x and y) are negative

So  that we can say it

V(b/p) = -28 × cos(45) i - 28 × sin(45) j     =  -19.8 i - 19.8 j  

and

now we use here relative motion  velocity for ball w.r.t ground

V(b/g) = V(b/p) + V(p/g )      ..........................2

put here value and we get

V(b/g) = (-19.8 i - 19.8 j) + 25 i     = 5.2 i - 19.8 j

so

Magnitude of that velocity

| V(b/g) | =  

| V(b/g) | = 20.47 m/sec

so that  Direction will be here

Direction = arctan (19.8 ÷ 5.2)

Direction = 75.3° South of East

so that

Speed = 20.47 m/sec at 75.3 deg South of East

and 2 significant  

Speed = 20 m/sec at 75 deg South of East = 20 m/sec at 15 deg East of South

electrical potential energy =0.072 j

explanation:

electrical potential energy is given by

u= q δv

q= charge=6 c

δv= potential difference and is given by

δv= e d

e= electric field

d= distance=5 mm=0.005 m

δv=2.4 (0.005)=0.012 v

now electric potential energy= u= q δv

u=6 (0.012)

u=0.072 j