Question

* A ball is projected horizontally from the top of

a building 19.6m high.
a, How long when the ball take to hit the ground?
b, If the line joining the point of projection to
the point where it hits the ground is 45
with the horizontal. What must be the
initial velocity of the ball?
c,with what vertical verocity does the ball strike
the grounds? (9= 9.8 M152)​

Answer 1

Explanation:

Given

Ball is projected horizontally from a building of height

time taken to reach ground is given by

(b) Line joining the point of projection and the point where it hits the ground makes an angle of

From the figure, it can be written

Considering horizontal motion

(c) The vertical velocity with which it strikes the ground is given by

Thus, the ball strikes with a vertical velocity of

Answer 2

Step-by-step explanation:

Given

Ball is projected horizontally from a building of height
`h=19.6\ m`

time taken to reach ground is given by

`\text{Cosidering vertical motion}\\\Rightarrow h=ut+0.5at^2\\\Rightarrow 19.6=0+0.5* 9.8t^2\\\Rightarrow t^2=4\\\Rightarrow t=2\ s`

(b) Line joining the point of projection and the point where it hits the ground makes an angle of
`45^(\circ)`

From the figure, it can be written

`\Rightarrow \tan 45^(\circ)=(h)/(x)\\\\\Rightarrow x=h\cdot 1\\\Rightarrow x=19.6`

Considering horizontal motion

`\Rightarrow x=u_xt\\\Rightarrow 19.6=u_x* 4\\\Rightarrow u_x=4.9\ m/s`

(c) The vertical velocity with which it strikes the ground is given by

`\Rightarrow v^2-u_y^2=2as\\\Rightarrow v^2-0=2* 9.8* 19.6\\\Rightarrow v=√(384.16)\\\Rightarrow v=19.6\ m/s`

Thus, the ball strikes with a vertical velocity of
`19.6\ m/s`