Question

In the figure here, a ball is thrown up onto a roof, landing 3.80 s later at height h = 25.0 m above the release level. The ball's path just before landing is angled at theta = 61.0 degree with the roof, (a) Find the horizontal distance d it travels. What are the (b) magnitude and (c) angle (relative to the horizontal) of the ball's initial velocity? Number Unit Number Unit Number Unit

Answer 1

just for simple calculation

+x in left direction

A. y0 = 25 m

y = 0 at t = 3.80 sec

y - y0 = Voy*t - 0.5*g*t^2

Voy = Vo*sin 61 deg

0 - 25 = Vo*3.80*sin 61 deg - 0.5*9.81*3.80^2

V = 13.78 m/sec

now in x-direction

x - xo = Vox*t

xo = 0 m

x - 0 = 13.78*3.80*cos 61 deg

x = 25.39 m

B.

Vx = Vox = 13.78*cos 61 deg = 6.68 m/sec

Vy = Voy - g*t

Vy = 13.78*sin 61 deg - 9.81*3.80 = -25.23 m/sec

|V| = sqrt (Vx^2 + Vy^2) =

|V| = sqrt (6.68^2 + 25.23^2)

|V| = 26.09 m/sec

C.

theta = arctan (Vy/Vx)

theta = arctan (-25.23/6.68) = - 75.17 deg

-ve sign means 75.17 deg from rightside.