Question

A stone is thrown from the top of a tower at an angle of 30 degree up with the horizontal with the Velocity of 16m/s after 4seconds of flight it strikes the ground calculate the height of the tower from the ground and the horizontal range of the stone

Answer 1

Answer:

The height of tower and horizontal range of the stone are 46.4 m and 55.4 m.

Explanation:

Given that,

Velocity =16 m/s

Time t =4 sec

Angle = 30°

We need to calculate the height of the tower

Using equation of projectile

$s_{y}=u_{y}t-\dfrac{1}{2}gt^2+h_{0}$

Where, $h_{0}$ = height

u =vertical velocity = $u\sin\theta$

t = time

Put the value in the equation

$0=16\times\dfrac{1}{2}\times4-\dfrac{1}{2}\times9.8\times16+h_{0}$

$-h_{0}=16\times\dfrac{1}{2}\times4-\dfrac{1}{2}\times9.8\times16$

$h_{0} = 46.4\ m$

Now, The horizontal range of the stone is

Using formula of horizontal range

$s_{x}=u_{x}t$

$s_{x}=16\cos30^{\circ}\times4$

$s_{x}=16\times\dfrac{\sqrt{3}}{2}\times4 m$

$s_{x}=55.4\ m$

Hence, The height of tower and horizontal range of the stone are 46.4 m and 55.4 m.

Answer 2

Answer: horizontal distance =55.4

Height of tower= 46.48

Explanation:

Initial velocity of stone u = horizontal component of 16 m/s

Ux = ucos theta =16 cos30 degree =16√3÷2= 8√3m/s

Vertical component

Uy= usin30 degree

16 sin30=8m/s

For height,

Sy=Uyt-1/2 gt^2+h

0=8×4+1/2×9.8×16+h

46.4=h

Horizontal component,

Sx=Uxt

=16 cos 30

=55.4