Question

a golfer hits a ball and gives it an initial velocity of 40m/s at an angle of 30 degrees above the horizonatal.
How long does the ball stay in the air?
How far horizontally (the range) does the ball travel before hitting the ground?

Answer 1

Answer:

Due to symmetry (you can verify this with calculation), the ball will be moving 80 ft/sec when it comes back down to the point from which you threw it on the building. So you can reduce the problem to describing a ball thrown off a building with initial velocity 80 ft/sec downward. We have

s(t)=32−80t−16t2,

as the ball's position function of time. To find t at s=0, we may use the quadratic formula, which gives

t=80±√802+4(16)(32)−32,

which simplifies to approximately t=0.37. Since acceleration is constant, we have

v(t)=−80−32t,

as the ball's velocity function of time. Solving for v when t=0.37, we obtain v=−91.84, which seems reasonable enough. The negative sign comes from calling the upwards direction + and the downwards direction −.