A projectile is thrown at an angle \beta with vertical .It reaches a maximum height H . The time taken to reach the highest point of its path is 

Answers (1)

@Harshit

As we have learned

Time of Flight -

Time for which projectile remains in the air above the horizontal plane.

- wherein

t=\frac{2U\sin \Theta }{g}

SO 

Time taken to reach highest point =\frac{t}{2}=\frac{U\sin \beta }{g}

 

 

 H = \frac{u^2\sin ^2 \beta }{2g} \; \; or \; \; (u\sin \beta )^2= 2gH

u\sin \beta = \sqrt{2gH}........(1)

Time taken to reach highest point = u\sin \beta/ g = \sqrt{2H/g}

Most Viewed Questions

Preparation Products

Knockout NEET 2024

Personalized AI Tutor and Adaptive Time Table, Self Study Material, Unlimited Mock Tests and Personalized Analysis Reports, 24x7 Doubt Chat Support,.

₹ 39999/-
Buy Now
Knockout NEET 2025

Personalized AI Tutor and Adaptive Time Table, Self Study Material, Unlimited Mock Tests and Personalized Analysis Reports, 24x7 Doubt Chat Support,.

₹ 45000/-
Buy Now
Knockout NEET 2023 (One Month)

Personalized AI Tutor and Adaptive Time Table, Self Study Material, Live Classes, Unlimited Mock Tests and Personalized Analysis Reports, 24x7 Doubt Chat Support,.

₹ 13999/- ₹ 6999/-
Buy Now
Knockout NEET (Three Months Subscription)

- AI Coach Study Modules, - Unlimited Mock Tests, - Expert Mentorship, - Study Improvement Plan.

₹ 5999/- ₹ 4999/-
Buy Now
Knockout NEET (Eight Months Subscription)

- AI Coach Study Modules, - Unlimited Mock Tests, - Expert Mentorship, - Study Improvement Plan.

₹ 15999/- ₹ 12499/-
Buy Now
Boost your Preparation for NEET 2021 with Personlized Coaching
 
Exams
Articles
Questions