Question

A projectile is thrown from the top of a building of height 43.0 m with an initial velocity 16.0 m/s along a direction that makes that makes an angle of 34.0 degrees above the horizontal.

(1) When does the projectile reach maximum height

(2) When does the projectile first reach a height of 45.0 m above the ground

(3) When does the projectile hit the ground and how far from the building

Answer 1

here,

the height of building , h0 = 43 m

initial velocity , u = 16 m/s

theta = 34 degree

a)

the time taken to reach the maximum height ,t1 = (u * sin(theta)) /g

t1 = ( 16 * sin(34)) /9.81 s

t1 = 0.91 s

b)

let the time taken be t2

for h = 45 m

using second equation of motion

h - h0 = u * sin(theta) * t2 - 0.5 * g * t2^2

45 - 43 = 16 * sin(34) * t2 - 0.5 * 9.81 * t2^2

solving for t2

t2 = 0.26 s

the time taken is 0.26 s

c)

let the time taken to hit the ground be t3

for vertical direction

using second equation of motion

- h0 = u * sin(theta) * t3 - 0.5 * g * t3^2

- 43 = 16 * sin(34) * t3 - 0.5 * 9.81 * t3^2

solving for t3

t3 = 4.01 s

the time taken is 4.01 s

the horizontal distance , x = u * cos(theta) * t3

x = 16 * cos(34) * 4.01 m

x = 53.2 m