A shot-putter throws the shot with an initial speed of 13 m/s at an angle of 42 degrees above the horizontal. Assuming that the shot is released from a height of 2.1 meters above the ground, calculate the horizontal distance the shot travels before it lands.
Answers
Answered by
1
Answer: 21.737 Meters
U y = 13 Sin 42 =8.67 m/ sec
at Max H , Vy = 0 so H additional (8.67)^2/10 = 7.51
T for additional hieght upwards = 8.67 / 10 = .867 Sec
So Noe the Shot put travel down from 2.1+7.51= 9.61 M
T down= Root ( 2*9.61/10)=Root ( 1.922) = 1.39 Sec
Total ime = 0.87+1.39=2.25 sec
DISTANCE in horizontal direction = 13 Cos 42 X 2.25 =21.737 Meters
Explanation:
Similar questions