Answer to Question #113126 in Mechanics | Relativity for Pierre Carlo Belarmino

Question

A projectile was launched from the top of a 25m high building at a velocity of +35m/s and at an angle of 28 degrees with the horizontal.

a)How far is the projectile from the building horizontally by the time it reaches its maximum height?

b) If the projectile lands on a hill 3 meters high from the ground, how long was the projectile airborne?

c) If the projectile lands on a hill 3 meters high from the ground,
how far is the hill from the building?

Answer 1

(a) the time needed for projectile to reach its maximum height

$$t=\frac{v_{iy}}{g}=\frac{v_i\sin\theta}{g}=\frac{35\sin 28^{\circ}}{9.8}=1.677\:\rm s.$$

The horizontal distance of the projectile at this instant

$$x=v_{ix}t=v_i\cos \theta t=35\times \cos 28^{\circ}\times 1.677=52\:\rm m$$

(b) the equation of motion of the projectile

$$y=y_0+v_{iy}t-\frac{gt^2}{2}$$

We get equation

$$3=25+35\sin 28^{\circ}t-\frac{9.8t^2}{2}$$

Root

$$t=4.4\:\rm s$$

(c) the hill is remote from the building by

$$x=v_{ix}t=v_i\cos \theta t=35\times \cos 28^{\circ}\times 4.4=136\:\rm m$$