Question

A body is thrown horizontally from a tower,
100 m high, at t = 0 with a velocity 10 m/s.
The velocity of the body is directed at an angle 45°
with the horizontal at a time equal to (g = 10 m/s2)
(1) 1s
(2) 2s
(3) 3 s
(4) 45

Answer 1

Answer:

45

Explanation:

i think this is the answer but not sure u refer any other book ect

Answer 2

Given that,

Height = 100 m

Initial velocity = 10 m/s

Angle = 45°

Acceleration due to gravity = 10 m/s²

The horizontal component of velocity is constant as no horizontal acceleration is present.

We need to calculate the velocity of projectile

Using formula of vertical component

$u=v\cos\theta$

$v=u\sec\theta$

Put the value into the formula

$v=10\sec45$

$v=14.14\ m/s$

We know,

Initial velocity along y axis

We need to calculate the final velocity along y axis

Using formula of velocity

$v_{y}=v\sin\theta$

Put the value into the formula

$v_{y}=14.14\sin45$

$v_{y}=10\ m/s$

We need to calculate the time

Using equation of motion

$v=u+at$

Put the value into the formula

$10=0+10\times t$

$t=1\ sec$

Hence, The velocity of the body is directed at an angle 45°  with the horizontal at a time is 1 sec.

(1) is correct option.