Question

a marble is to be thrown horizontally from a height of 19.6 cm above the ground so that it hits another marble on the ground,2 m away.the velocity of throw should be?

Answer 1

this question is based on horizontal projectile.

we know, trajectory of horizontal projectile is given by, $\boxed{y=\frac{gx^2}{2u^2}}$

where y is vertical distance . e.g., height of particle from the ground.

x is horizontal distance, g is acceleration due to gravity and u is initial velocity of particle.

here, y = 19.6cm = 0.196m, x = 2m

and g = 9.8 m/s² [ experimental value ]

so, y = gx²/2u²

0.196 = 9.8 × (2)²/2u²

or, 0.196 = 9.8 × 2/u²

or, 0.196 = 19.6/u²

or, 100 = u² => u = 10

hence, velocity of thrown should be 10m/s

Answer 2

Answer:

Explanation:

Hopes u understand