Question

Consider a rock thrown off a bridge of height 79.8 m at an angle theta = 25 degree with respect to the horizontal as shown in Figure P4.20. The initial speed of the rock is 10.5 m/s. Find the following quantities: the maximum height reached by the rock the time it takes the rock to reach its maximum height the time at which the rock lands how far horizontally from the bridge the rock lands the velocity of the rock (magnitude and direction) just before it lands, magnitude direction: Give your answer in degrees. If you think the answer is "20 degrees down from the positive x-axis", you would enter "-20" (note the negative sign!)

Answer 1

Given data is bridge height h=79.8m,angle $\theta$ =25,initial velocity u=10.5m/s

a)the maximum height h=u²sin² $\theta$ /2g=(10.5)²sin²25/2*9.8=1.004m

b)time to reach rock max height t=usin $\theta$ /g=10.5*0.4226/9.8=0.453sec

c)total time to reach land h=-usin $\theta$ +(gt²/2)

79.8=-10.5*sin25*t+(9.8*t²/2)

solve this equation t=4.5sec

d)the horizontal distance cover bythe rock R=u²sin2 $\theta$ /g

R=10.5²*sin50/9.8=8.6175m

e)final velocity of rock just before reach the ground is v=usin $\theta$ -gt

v=10.5*sin25-9.8*4.5=-39.6626m/s negative direction