Question

A particle is thrown vertically with a velocity 50ms^2.How high would the particle rise and what time would it take to reach the highest point? (g=10 ms ^2) ​

Answer 1

Explanation:

GIVEN:

A PARTICLE IS THROWN VERTICALLY WITH INITIAL VELOCITY OF 50 M /S

g=10 m/s^2

TO FIND:

• MAXIMUM HEIGHT

•TIME TAKEN.

CONCEPT:

IF ANY PARTICLE IS THROWN VERTICALLY UP THEN IT'S FINAL VELOCITY BECOMES 0 M/S WHEN ITS REACHES AT THE MAXIMUM HEIGHT.

FORMULAS TO BE USED:

•V^2=U^2-2gs

•V=U-gt

SOLUTION:

FINDING THE MAXIMUM HEIGHT:

=>V^2=U^2-2gs

=>0=50^2-2×10×S

=>0=2500-20S

=>-2500=-20S

=>S=-2500/-20=125 metre.

NOW FINDING THE TIME TAKEN:

=>V=U-gt

=>0=50-10t

=>-50= -10t.

=>t=-50/-10=5 seconds.