
A body is thrown with a velocity of 9.8m/s making an angle of ${30^ \circ }$ with the horizontal. It will hit the ground after a time:
A) 1.5 s
B) 1 s
C) 3 s
D) 2 s
Answer
123.6k+ views
Hint: The motion of an object thrown or projected into the air, which is a subject to only the acceleration of gravity is known as projectile motion. The object is called a projectile, and the path covered by the object or projectile is called its trajectory.
Formula Used: The numerical formula to calculate the time of flight of a projectile is given by the mathematical expression given below:
\[T = \dfrac{{2 \times {u_y}}}{g}\]
In this mathematical equation to calculate the time of flight of a projectile, ${u_y}$ is the vertical velocity of the projectile or the velocity of the projectile along the direction of the y-axis.
$g$ is the acceleration due to gravity which is numerically equal to $9.8m{s^{ - 2}}$.
Complete step by step answer:
One of the most important things illustrated by projectile motion is that vertical and horizontal motions are independent of each other. Galileo was the first person to fully comprehend this characteristic. He used it to predict the range of a projectile.
In this question we have to take into consideration the velocity of the projectile along the y-axis or the vertical velocity of the object or projectile. Thus, we have:
${u_y} = u\sin \theta $
Now we substitute this equation in the mathematical formula to obtain the final expression for time of flight. Thus:
$T = \dfrac{{2 \times u\sin \theta }}{g}$
Now, we know that from the data given in the numerical problem.
$u = 9.8m/s$, $\theta = 3{0^ \circ }$ and $g = 9.8m{s^{ - 2}}$
Thus, substituting these values in the above mathematical equation, we get:
$T = \dfrac{{2 \times 9.8 \times \sin 30^\circ }}{{9.8}}$
Solving this equation further, we obtain:
$ \Rightarrow T = 2 \times \dfrac{1}{2} = 1s$
Thus the projectile takes one second to hit the ground.
Note: Note that this definition assumes that the upwards direction is defined as the positive direction. If we arrange the coordinate system instead such that the downwards direction is positive, then acceleration due to gravity takes a positive value.
Formula Used: The numerical formula to calculate the time of flight of a projectile is given by the mathematical expression given below:
\[T = \dfrac{{2 \times {u_y}}}{g}\]
In this mathematical equation to calculate the time of flight of a projectile, ${u_y}$ is the vertical velocity of the projectile or the velocity of the projectile along the direction of the y-axis.
$g$ is the acceleration due to gravity which is numerically equal to $9.8m{s^{ - 2}}$.
Complete step by step answer:
One of the most important things illustrated by projectile motion is that vertical and horizontal motions are independent of each other. Galileo was the first person to fully comprehend this characteristic. He used it to predict the range of a projectile.
In this question we have to take into consideration the velocity of the projectile along the y-axis or the vertical velocity of the object or projectile. Thus, we have:
${u_y} = u\sin \theta $
Now we substitute this equation in the mathematical formula to obtain the final expression for time of flight. Thus:
$T = \dfrac{{2 \times u\sin \theta }}{g}$
Now, we know that from the data given in the numerical problem.
$u = 9.8m/s$, $\theta = 3{0^ \circ }$ and $g = 9.8m{s^{ - 2}}$
Thus, substituting these values in the above mathematical equation, we get:
$T = \dfrac{{2 \times 9.8 \times \sin 30^\circ }}{{9.8}}$
Solving this equation further, we obtain:
$ \Rightarrow T = 2 \times \dfrac{1}{2} = 1s$
Thus the projectile takes one second to hit the ground.
Note: Note that this definition assumes that the upwards direction is defined as the positive direction. If we arrange the coordinate system instead such that the downwards direction is positive, then acceleration due to gravity takes a positive value.
Recently Updated Pages
Difference Between Circuit Switching and Packet Switching

Difference Between Mass and Weight

JEE Main Participating Colleges 2024 - A Complete List of Top Colleges

JEE Main Maths Paper Pattern 2025 – Marking, Sections & Tips

Sign up for JEE Main 2025 Live Classes - Vedantu

JEE Main 2025 Helpline Numbers - Center Contact, Phone Number, Address

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility & More

JEE Main Login 2045: Step-by-Step Instructions and Details

Class 11 JEE Main Physics Mock Test 2025

JEE Main Exam Marking Scheme: Detailed Breakdown of Marks and Negative Marking

JEE Main 2023 January 24 Shift 2 Question Paper with Answer Keys & Solutions

JEE Main 2025: Conversion of Galvanometer Into Ammeter And Voltmeter in Physics

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

NCERT Solutions for Class 11 Physics Chapter 1 Units and Measurements

NCERT Solutions for Class 11 Physics Chapter 9 Mechanical Properties of Fluids

Units and Measurements Class 11 Notes: CBSE Physics Chapter 1

JEE Advanced 2025: Dates, Registration, Syllabus, Eligibility Criteria and More

NCERT Solutions for Class 11 Physics Chapter 2 Motion In A Straight Line
