A particle is thrown vertically with a velocity of 50m/s. What time would it take to reach the height point? What would be its velocity at highest point? How high would the particle rise?

Open in App

Solution

U= 50m/s

g= -10m/s²

The velocity at the highest point will be 0 m/s
Thus v=0

Height can be calculated by

2gs=v² - u²
-20×s= -2500
s= 125m

Time :
v= u+gt
0=50-10t
-50=-10t
t= 5sec

Suggest Corrections

Similar questions

Q. A particle is thrown vertically with a velocity of

50 m / s

. What will be its velocity at the highest point of the journey? How high would the particle rise? What time would it take to reach the highest point ? (

g = 10 m / s^{2}

)

A particle is thrown upward with a velocity of 50m/sec .what will be the velocity at its highest point of the journey. How high the particle rise. What time would it taken to reach the highest point (g=10m/sec)

Q. A particle is dropped from a

200 m

tall tower. At the same time, a second particle is thrown upwards with velocity

50 m / s .

Calculate when and where would these two particles meet?

Q. A ball thrown in vertical upward direction attains maximum height of

16 m

. At what height would its velocity be half of its initial velocity?

Q. A particle is projected with a velocity

u,

at an angle

α,

with the horizontal. At what time its vertical component of velocity becomes half of its net speed at the highest point will be

Join BYJU'S Learning Program

A particle is thrown vertically with a velocity of 50m/s. What time would it take to reach the height point? What would be its velocity at highest point? How high would the particle rise?

U= 50m/s g= -10m/s² The velocity at the highest point will be 0 m/s Thus v=0 Height can be calculated by 2gs=v² - u² -20×s= -2500 s= 125m Time : v= u+gt 0=50-10t -50=-10t t= 5sec

U= 50m/s

g= -10m/s²

The velocity at the highest point will be 0 m/s
Thus v=0

Height can be calculated by

2gs=v² - u²
-20×s= -2500
s= 125m

Time :
v= u+gt
0=50-10t
-50=-10t
t= 5sec