Question

A body is projected with an initial velocity of 20m/s at an angle of 30° with the horizontal. calculate (a) maximum height (b) time taken to reach the maximum height (c) horizontal range​

Answer 1

Answer:

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Answer 2

Answer: The answer to the given question is:-

a) Maximum height= 5 m

b) Time taken to reach maximum height= 1-second

c) Horizontal range= 34.64 m

Explanation:

The body is undergoing a projectile motion i.e. it is projected with an initial velocity of 20 m/sec at an angle of 30 degrees with the horizontal.

From this, we can say the angle made with vertical is 90-30= 60 degrees.

a) maximum height is the topmost point, where velocity along y-direction (Vy) becomes zero.

$2gh= (v_y)^2 -(u_y)^2\\2(-10)h= 0-(20sin30)^2\\h=\frac{-(20sin30)^2}{-20}=5 m$

b)we can use the 1st equation of motion for this, as  $v_y=0$ and $u_y= 20 sin 30= 10 m/sec$ and $g= -10 m/sec^2$

$v_y= u_y + gt\\ 0=10-10t\\t= \frac{10}{10}=1 sec$

c) Range is the displacement in the horizontal direction. There is no acceleration in the horizontal direction. So it works like the case of uniform motion.

$R= u_ycos30 \times \frac{2u_ysin30}{g}= 20\times\frac{\sqrt{3}}{2} \times40\times\frac{1}{2}=34.64 m$