Question

a stone thrown horizontally with speed 30 m/s from a tower of height 50 m .the height of stone from ground at which it makes 45degrees with horizontal is

Answer 1

Let the height of stone is h from the group at which it makes 45° with horizontal.
horizontal initial velocity = horizontal final velocity
30 = ucos45°
u = 30/cos45° = 30√2 m/s ........(i)

for vertical motion,
$v_y^2=u_y^2+2g(50-h)$
u²sin²45° = 0 + 2g(50-h) [ initially stone is thrown horizontally so, vertical initial velocity = 0]
put equation (i),
(30√2)² sin²45° = 2× 10 × (50-h)
900 × 2 × 1/2 = 20(50-h)
900 = 20(50-h)
50 - h = 45
h = 5m

hence, answer should be 5m

Answer 2

We know that whenever any object is thrown from certain height and it falls freely under the gravity then it is known as Projectile. It means it will follow the Parabolic Paths.

Given conditions ⇒

Height of the Tower = 50 m.

Speed of the Stone in x - direction = 30 m/s.

This is the component of the speed of the in x-direction.

The component of the x direction will remain the same throughout the motion where as the component in y-direction will continue to increase.

Now, See the Figure attached for understanding it more carefully.

From the Figure,

Component in Horizontal direction = u Cosθ

30 = u Cos 45°

⇒ u = 30√2 m/s.

Component in Vertical Direction = u Sin θ

= 30√2 × Sin 45° = 30 m/s.

Now, Velocity in y direction  at the point where angle is noticed is  30 m/s.

This means it is the Final Velocity at that time.

∴ v =  30 m/s.

u = 0 [Since, Initial velocity of y-direction will be zero)

Also, Height of the body at that point where angle is noticed is a meter.

Now,  Total height covered by the object = 50 - a.

Using the Velocity- Position Relation,

v² - u² = 2gS

(30)² = 2 × 10 × (50 - a)

⇒ 900/20 = 50 - a

a = 50 - 45

∴ a = 5 m.

Hence the height of the body where angle is formed is 5 m.

Hope it helps.