Question

YOU THROW A MARBLE HORIZONTALLY FROM A TABLE TOP WITH A VELOCITY OF 1.50 M/S. THE MARBLE FALLS 0.70 AWAY FROM THE TABLES EDGE

Answer 1

Answer:

Gives

Velocity (v0x)=1.50m/sec

d=0.70m

We know that

d=V_{0x}td=V  

0x

​

t

0.70=1.50\times t0.70=1.50×t

t=0.47 sec

V_{fy}=V_{0y}+gtV  

fy

​

=V  

0y

​

+gt

V_{fy}=0+(-9.8)\times0.47V  

fy

​

=0+(−9.8)×0.47

V_{fy}=-4.5783m/secV  

fy

​

=−4.5783m/sec

V=\sqrt{V_{ox}^2+V_{fy}^2}V=  

V  

ox

2

​

+V  

fy

2

​

 

​

 

Put value

V=\sqrt{{1.50}^2+{(-0.47)}^2}V=  

1.50  

2

+(−0.47)  

2

 

​

 

V=4.82m/sec

Explanation:

I hope you like my answer so please mark it as brainliest and give thanks

Answer 2

Your complete question is - A marble is thrown horizontally from a table top with a velocity of 1.50 m/s. The marble fall 0.70 m away from the table's edge. How high is the table? What is the marble's velocity just before it hits the floor?

solution : it is based on horizontal projectile motion.

we know, trajectory of horizontal projectile motion is given by, y = gx²/2u²

here given, x = 0.7 m, u = 1.5 m , g = 9.8 m/s²

so , y = (9.8)(0.7)²/(2 × 1.5²)

= 1.0671 m

Therefore the table is 1.0671 m high from the ground.

we know, horizontal velocity remains constant because there is no acceleration acting in horizontal direction.

using formula, s = ut + 1/2 at²

for horizontal direction, a = 0

s = x = 1.5 t

⇒0.7 = 1.5 t

⇒t = 7/15 sec

but in vertical direction, acceleration due to gravity acts on it.

so, velocity in horizontal direction,

v = u + at

= 1.5 + (-9.8)(7/15)

= 1.5 - 4.57

= -3.07 m/s

Therefore velocity just before it strikes, v = 1.5 i - 3.07 j