Question

A ball is projected at an angle of 30 degree above with the horizontal from top of a tower and strikes the ground in 5 second at an angle of 45 degree with the horizontal find the height of the tower and the speed with which it was projected

Answer 1

Final Answer : H = 125 (2-√3)m
|u| = 50(√3-1) m/s .

Steps and Understanding:
1) Let the initial velocity of projectile be u m/s at an angle of 30° from horizontal on top of a tower of height H.
For figure see pic.

2) Given,
Initial velocity, u = u (say)
Height of tower, h = H
Projection angle,theta = 30°
Fall at angle 45° with the horizontal on ground.
a(y) = -10m/s^2 (g)
a(x) = 0
time taken to reach ground, t = 5s
Vertical upward direction is +.
and vice versa.

3) Vertical Equation of motion,
$s(y) = u(y)t \: + \frac{1}{2} a(y) {t}^{2} \\ - h = u \sin(30) \times 5 - \frac{1}{2} \times 10 \times {5}^{2} \\ = > - h = \frac{5u}{2} - 125 \: \: \: - - - (1)$

4)
$v(y) = u(y) + a(y)t \\ = > v(y) = u \sin(30) - 10 \times 5 \\ = > v(y) = \frac{u}{2} - 50$

5) Since, v(y) is negative as v(y) is in downward direction, so
$|v(y)| = 50 - \frac{u}{2}$
Since, F(x) =0
so, velocity along horizontal direction will be constant.
v(x) = u cos(30°)
$|v(x)| = \frac{u \sqrt{3} }{2}$

And looking at pic,
$\tan(45) = \frac{ |v(y)| }{ |v(x)| } \\ = > |v(y)| = |v(x)| \\ = > 50 - \frac{u}{2} = \frac{u \sqrt{3} }{2} \\ = > u = \frac{50( \sqrt{3} + 1)}{2} \\ = > u = 50( \sqrt{3} - 1)$

6) Using u in equation (1) ,
$h = 125 - \frac{5u}{2} \\ = > h = 125 - \frac{5 \times 50( \sqrt{3} - 1)}{2} \\ = > h = 125(2 - \sqrt{3} )m$

Therefore,
speed of projection u = 50(√3-1)m/s
Height of tower, H = 125(2-√3)m

Answer 2

Answer:

Final Answer : H = 125 (2-√3)m

|u| = 50(√3-1) m/s .

Steps and Understanding:

1) Let the initial velocity of projectile be u m/s at an angle of 30° from horizontal on top of a tower of height H.

For figure see pic.

2) Given,

Initial velocity, u = u (say)

Height of tower, h = H

Projection angle,theta = 30°

Fall at angle 45° with the horizontal on ground.

a(y) = -10m/s^2 (g)

a(x) = 0

time taken to reach ground, t = 5s

Vertical upward direction is +.

and vice versa.

3) Vertical Equation of motion,

4)

5) Since, v(y) is negative as v(y) is in downward direction, so

Since, F(x) =0

so, velocity along horizontal direction will be constant.

v(x) = u cos(30°)

And looking at pic,

6) Using u in equation (1) ,

Therefore,

speed of projection u = 50(√3-1)m/s

Height of tower, H = 125(2-√3)m

Step-by-step explanation: