Question

A ball is thrown upward at an angle of 30° with the horizontal and lands on the top edge of a building
that is 20 m away. The top edge is 5m above the throwing point. The initial speed of the ball in
metre/second is (take g=10 m/s2):
(A) U - 40 114+3)
V 13.13
(B) u = 40
V13 do mis
40
(C)u-40, 4+ mis
(D) u = 40
m/s
v 13
13 (4+13)​

Answer 1

Explanation:

go through the attachment.

hope, it helped you

Answer 2

Answer:

The initial speed of the ball is 20 m/s.

Explanation:

Given that,

Horizontal angle = 30°

Horizontal distance = 20 m

Height = 5 m

We need to calculate the initial velocity

Using equation of trajectory

$y= x\tan\theta+\dfrac{gx^2}{2u^2\cos^2\theta}$

Where,

y = vertical distance

x = horizontal distance

u = initial velocity

g = acceleration due to gravity

Put the value into the formula

$5=20\times\tan30^{\circ}+\dfrac{10\times20^2}{2u^2\cos^230^{\circ}}$

$5=20\times0.57+\dfrac{4000}{2u^2\times\dfrac{3}{4}}$

$5-20\times0.57=\dfrac{8000}{3u^2}$

$u=20\ m/s$

Hence, The initial speed of the ball is 20 m/s.