Question

A lab cart flies off the end of a track on a lab table with a horizontal velocity of 2.08 m/s. If it lands 0.96 meters from the edge of the table, how tall is the table?

Answer 1

Answer :-

The table is 1.037 metres tall .

Explanation :-

We have :-

→ Horizontal velocity (v) = 2.08 m/s

→ Landing distance (d) = 0.96 m

→ Gravitational acceleration (g) = 9.8 m/s²

________________________________

Firstly, we shall calculate time taken by the lab cart to cover the given distance .

⇒ Time = Distance/Speed

⇒ t = d/v

⇒ t = 0.96/2.08

⇒ t ≈ 0.46 s

We have got the time of fall of the lab cart . Now, we can calculate height of the table by substituting values in the 2nd equation of free-fall . [u = 0 m/s] .

h = ut + ½gt²

⇒ h = 0(0.46) + ½ × 9.8 × (0.46)²

⇒ h = 0 + 4.9 × 0.2116

⇒ h ≈ 1.037 m

Answer 2

To Find :

The hieght of the table , when the horizontal velocity is given

__________________________

Given :

• Velocity = 2.08 m/s
• Distance of landing = 0.96 m
• here, Acceleration = gravity
• gravity = 9.8 m/s²

__________________________

Formula to be used :

• Time = distance/velocity
• S = ut+½at²

here, s is the height of the table and acceleration is the gravity , hence

• h = ut+½gt²

__________________________

Solution :

Given, Velocity = 2.08 m/s

Distance = 0.96 m

so, time = 0.96/2.08

or, time ≈ 0.462

Given, acceleration = 9.8 m/s²

initial velocity = 0.96 m/s

time = 0.462

g = 9.8 m/s²

$\tt{ \implies \: h = (0.96 \times 0.462) + \frac{1}{2} \times 9.8 \times 0.462}$

$\implies \tt{h = 0.443 + \frac{1}{2} \times 2.05}$

$\implies \tt{h \approx 0.444 + 1.025}$

$\implies \tt{h \approx \: 1.467}$

Hence, height ≈ 1.467 m