Question

A bomb is fired from a cannon with a velocity of 1000m/s making an angle 30 with the horizontal. what is the time taken by the bomb to reach the highest point

Answer 1

Velocity of bomb is 1000 m/s at angle of 30 degree with the horizontal

now we can have its velocity components along X and Y directions

$v_y = vsin30 = 1000*0.5 = 500 m/s$

$v_x = vcos30 = 1000*0.866 = 866 m/s$

now in order to find the time to reach the highest point we can use kinematics equation

as the object will reach at highest point its vertical component of velocity will become zero

so we can use

$v_f = v_i + at$

now we can plug in all value for y direction

$0 = 500 - 10 * t$

here acceleration in vertical direction is due to gravity so it is a = -10 m/s^2

now by solving above equation

$0 = 500 - 10 t$

$t = 50 s$

so it will take 50 s to reach the top.

Answer 2

Answer:

50s

Explanation:

Velocity of bomb υ=1000 m/s

The angle θ=30∘

Time taken by the bomb to reach the highest point

t=υsinθg

=1000sin30x9.8

So,

t=1000×0.59.8

=50sec