Question
A golfer launches a golf ball off a tee that has been raised 24.8 meters off the ground. If the golfer strikes the ball with a velocity of 26.5 m/s at an angle of 45.8 degrees, how far will the ball carry in the air before hitting the ground below?
Answers
First solve for the time t that the golf ball is in the air.
24.8 + 26.5sin45.8*t - 4.9*t^2 = 0
Use the quadratic equation and take the positive root for t.
The horizontal X distance travelled is
X= 26.5*cos45.8*t
24.8 + 26.5sin45.8*t - 4.9*t^2 = 0
Use the quadratic equation and take the positive root for t.
The horizontal X distance travelled is
X= 26.5*cos45.8*t
physics is stupid
Related Questions
A golfer hits a golf ball with a velocity of 36.0 meters/second at an angle of 28.0°. If the hang ti...
A golfer hits a golf ball, giving it an initial speed of 39.7 m/s at an angle of 55° above the horiz...
a golfer hits a golf ball an initial velocity of 54m/s [42 degrees above the horizontal]. the ball l...