Question

A solid metal ball is thrown from the top of a building at an angle of 35° above the horizontal with an initial speed of 15 m/s. The ball lands on the ground 2.7 s after it is thrown. What is the height of the building (in m)?

Answer 1

Let us consider the height of the building = H
Now initial velocity in the Y direction (u) = VSin$\theta$ = 15Sin35 = 8.604 m/s
time taken to reach the ground (t) = 2.7 s
Considering upward direction as positive therefore the displacement in the Y direction
S = -H
Now using kinematic equation , we can write
S = ut +(1/2)at²
where u is initial velocity in y direction
Now putting all the values
$-H = 8.604*2.7 + (0.5*(-9.81)2.7^{2})$
H = 12.53 m
Hence the height of the building will be 12.53 m

sin35 9- 35