Question

11. A golfer hits a 42 g ball, which comes down on a tree root and bounces straight up with an initial speed of 15.6 m/s. Determine the height the ball will rise after the bounce. Show all your work.

Answer 1

Explanation:

Given:-

• Mass of ball ,m = 42g

• Initial velocity ,u = 15.6m/s

• Final velocity ,v = 0m/s

• Acceleration due to gravity ,g = 9.8m/s

To Find:-

• Height ,h

Solution:-

We have to calculate the height attained by the ball after bounce. Using 3rd Equation of Motion

• v² = u² +2gh

v is the final velocity

g is the acceleration

u is the initial velocity

h is the height attained

Substitute the value we get

→ 0² = 15.6² + 2×(-9.8) × h

→ 0 = 243.36 + (-19.6) ×h

→ -243.36 = -19.6×h

→ h = -243.36/-19.6

→ h = 243.36/19.6

→ h = 12. 41 m

Therefore, the height attained by the ball is 12.41 Metres.

Answer 2

♣ɢɪᴠᴇɴ:–

• Mass of the ball (m) = 42 g

• Initial velocity (u) = 15.6 m/s

• Final velocity (v) = 0 m/s

• Acceleration (a) = 9.8 m/s²

♣ᴛᴏ ғɪɴᴅ:–

• The height attained by the ball (h) = ?

♣sᴏʟᴜᴛɪᴏɴ:–

Using the third equation of motion we can easily calculate the height attained by the ball :–

• v² = u² + 2ah

→ 0² = 15.6² + 2 × (-9.8) × h

→ 0 = 243.36 + (-19.6h)

→ 0 - 243.36 = -19.6h

→ -243.36 = -19.6h

→ h = -243.36/-19.6

→ h = 12.41 m

♣ᴀɴsᴡᴇʀ:–

• The height attained by the ball is 12.41 m