Question

A projectile is thrown from a point in a horizontal plane
such that its horizontal and vertical velocity component
are 9.8 m/s and 19.6 m/s respectively. its horizontal
range is :-
(1) 4.9 m (2) 9.8 m (3) 19.6 m (4) 39.2 m​

Answer 1

Answer:

39.2m

Explanation:

y=ut+1/2at square

o=19.6t -1/2gt sq.

t=4s

then ,X=ut +1/2at sq.

R= 9.8×4 +0 =39.2 m

Answer 2

ANSWER:

39.2 m

EXPLANATION:

The range of the projectile thrown at a distance on a horizontal plane formula is  

$R=\frac{2 u \sin \theta u \cos \theta}{g}$

The horizontal velocity of the range is 9.8 m/s

The vertical velocity of the range is 19.6 m/s

Now as we all know that the horizontal equation of R or x = usinθ  

And the value of the vertical y of the Range, R is y = ucosθ

Therefore, the horizontal value of the Range, R  

$R=\frac{2 u \sin \theta . u \cos \theta}{g}$

$R=\frac{9.8 \times 19.6 \times 2}{9.8}=39.2$

Therefore, the horizontal range R = 39.2 m