The figure shows the trajectory of a golf ball in projectile motion over level g

Question

The figure shows the trajectory of a golf ball in projectile motion over level ground. The time t0 = 0 s corresponds to the moment just after the ball is launched from position x0 = 0 m, y0 = 0 m. Its initial velocity (launch velocity) is v0. Two other points along the trajectory are indicated in the figure. At time t1 (with velocity v1), the ball reaches the peak of its trajectory. Its position at this moment is denoted by (x1,y1) = (x1,ymax), since it is at its maximum height. At time t2 (with velocity v2), just before the ball touches the ground again, its position is (x2,y2) = (xmax,y2), since it is at its maximum horizontal range.

2. [1pt] Assume an initial speed of 30.9 m/s, at an angle of 18.2

Explanation / Answer

1)ACEH

2)vx=30.9cos(18.2)=29.35m/s

vy=30.9Sin(18.2)=9.65m/s

3)ax=0

ay=-9.81m/s^2

4)v1x=30.9cos(18.2)=29.35m/s

v1y=0

5)ax=0

ay=-9.81m/s^2

6)t=2*vy/g=2*9.65/9.81=1.97s

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.