Agolfer tees off and hits a golf ball at a speed of 31 m/s and at an angle of 35 degrees. how far did the ball travel before hitting the ground? round the answer to the nearest tenth of a meter.

92.1 meters

Explanation:

Given:

x₀ = 0 m

y = y₀ = 0 m

v₀ₓ = 31 cos 35° m/s

v₀ᵧ = 31 sin 35° m/s

aₓ = 0 m/s²

aᵧ = -9.8 m/s²

Find: x

First, we need to find the time:

y = y₀ + v₀ t + ½ at²

0 = 0 + (31 sin 35) t − 4.9 t²

t = 0 or (31 sin 35) / 4.9

Now find x:

x = x₀ + v₀ t + ½ at²

x = 0 + (31 cos 35°) (31 sin 35) / 4.9 + 0

x ≈ 92.1

The ball travels 92.1 meters.

diagram is missing, but i will try to explain your doubt.

answer:

amplitude is the maximum displacement or distance moved on a point of a vibrating body or wave measured from its equilibrium position.it is equal to half the length of vibration path.