Question

10.) A ball is thrown from the top of a tower of height
63.8 m with a velocity of 30 ms-1 at an angle of 37°
above the horizontal. The angle formed by the
velocity of the stone with the horizontal when it hits
the ground is (take g = 10 m/s2)​

Answer 1

A ball is thrown from the top of a tower of height 63.8 m with a velocity of 30m/s at an angle of 37° above the horizontal.

time taken to reach the ground,

-63.8 = (30sin37°)t - 1/2 × 10 × t²

⇒-63.8 = 18t - 5t²

⇒5t² - 18t - 63.8 = 0

⇒t ≈ 5.8 sec

x - component : initial velocity , u = ucos37°

= 30 × 4/5 = 24 m/s

velocity after t sec, v = u + at

= 24 + 0 × 5.8 = 24 m/s [ acceleration along horizontal direction ]

y - component : initial velocity , u'= usin37° = 30sin37° = 18 m/s

velocity after t sec, v' = u' + a't

= 18 + (-10)(5.8)

= 18 - 58 = -40 m/s

so, velocity after t sec in vector form, v = 24 i - 40 j

angle formed by the velocity of the stone with the horizontal when it hits the ground is θ = tan-¹(40/24) = tan-¹(5/3) ≈ 59.04°

Answer 2

$\huge\bold\purple{Answer:-}$

A ball is thrown from the top of a tower of height 63.8 m with a velocity of 30m/s at an angle of 37° above the horizontal.

time taken to reach the ground,

-63.8 = (30sin37°)t - 1/2 × 10 × t²

⇒-63.8 = 18t - 5t²

⇒5t² - 18t - 63.8 = 0

⇒t ≈ 5.8 sec

x - component : initial velocity , u = ucos37°

= 30 × 4/5 = 24 m/s

velocity after t sec, v = u + at

= 24 + 0 × 5.8 = 24 m/s [ acceleration along horizontal direction ]

y - component : initial velocity , u'= usin37° = 30sin37° = 18 m/s

velocity after t sec, v' = u' + a't

= 18 + (-10)(5.8)

= 18 - 58 = -40 m/s

so, velocity after t sec in vector form, v = 24 i - 40 j

angle formed by the velocity of the stone with the horizontal when it hits the ground is θ = tan-¹(40/24) = tan-¹(5/3) ≈ 59.04°