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Grade 11Mechanics

Ball is thrown upward at an angle of 30° with the horizontal and lands on the top edge of a building that is 20 m away the top touches 5 m above the throwing point the national speed of the ball in metre per second is

Aditya Jaiswal

7 Years agoGrade 11

2 Answers

APARNA SHAJI

7 Years ago

here we have two velocities one in the horizontal and othrt in the vertical direction so for the velocity of the particle you have to take the vector sum of it. H=U^2SIN^2 30/2g --(1) (vertiical)and R = U^2 SIN2*30 /g--(2)(horizontal).so you get U^2 ALONG VERTICAL DIRECTION AS 400 AND U^2 ALONG HORIZONTAL DIRECTION AS 400/ROOT3. SO BY GETTING THE ROOT MEAN SQUARE OF BOTH U GET THE ANSWER AS APPROX 600m/s

Khimraj

7 Years ago

Horizontal Component

The horizontal component of the velocity is constant as it is perpendicular to g.

So we can write:

vcos30=20t

Where t is the time of flight.

∴v×0.866=20t

t=200.866v (1)

Vertical Component

We can use:

s=ut+12at2

This becomes:

5=vsin30t−12gt2

∴5=v×0.5×t−12×9.8×t2

5=v×0.5×t−4.9t2 (2)

We can substitute the value of t from (1) into the first part of (2)⇒

∴5=v×0.5×200.866v−4.9t2

∴5=100.866−4.9t2

4.9t2=11.547−5=6.547

4.9t2=6.547

t2=6.5474.9=1.336

t=√1.336=1.156xs

We can now put this value of t back into (1)⇒

200.866v=t

∴0.866v=20t=201.156

∴v=201.156×0.866=19.97xm/s

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