Word problem, average velocity?

Question

An observer on a golf course, at 2:00 in the afternoon, stands 60 m west of a player who drives a ball due north down the fairway. If the ball lands 2.0 seconds later, 156 m from the observer, what was its average velocity?

Answer 1

Create a right triangle.

Let the opposite be a.

Let the adjacent be b.

Let the hypotenuse be c.

Given b = 60m and  c = 156m.

Using pythagoras' theorem,

a² + b² = c²

a = 144m

The ball flew 144m in 2s, therefore the average velocity is 72m/s.

Answer 2

yes consider its a right triangle.

if the man is 60 m away from the golfer let 60m be the base. and 156m away from the observer be r. u need to find the distance of the ball from the golfer.

first find the angle data 156cosdata=60m, data=cos^-1(60/156)=67.4 deg

then that means 156sin(67.4)= 144m this is the distance from the golfer to the ball

to find average Velocity ,  v= displacement over time, so

average v=144m/2s=72m/s

Answer 3

Consider a right triangle. Let a side a be the side that runs due north. Let a side b = 60 * m be the side that runs west. Let the hypotenuse c = 156 * m be the distance from the ball to the observer. Find a. Use the theorem of Pythagoras; thus, a = sqrt[(c^2)-(-b)^2] = sqrt[24,336 * (m^ 2)  - 3,600 * (m^2)] = sqrt[20,736 * (m^2)] = 144 * m.

So, the average velocity v = a/t = 144 * m / (2.0 * s) = 72 * m / s.