Question

A projectile is projected with initial speed 10 m/s
at an angle 37° with the horizontal. The speed of
the projectile at its maximum height will be
(1) 10 m/s
(2) 4 m/s
(3) 6 m/s
(4) 8 m/s​

Answer 1

At highest point vertical component of velocity is zero
There is only horizontal component = vcos 37
=10 cos 37
=10(4/5)
=8 m/s

Answer 2

The speed at the maximum height is (4)8 m/sec.

Given,

Initial speed of the projectile=10 m/s

Angle at which it was thrown=37° with the horizontal.

To find,

the speed of the projectile at its maximum height.

Solution:

• The motion here is a projectile motion.
• When a projectile is thrown with a velocity u, it has two components usinθ on the y-axis and ucosθ on thex-axis.
• When the object is at its maximum height usinθ becomes 0.
• It is only ucosθ which remains constant for the whole motion.
• The maximum height attained by a projectile is given as:
• $H=\frac{u^{2}sin^{2}(theta) }{2g}$.
• where, H-maximum height, u-initial speed, theta-angle with the horizontal at which the projectile was thrown and g-acceleration due to gravity.

The speed of the projectile at its maximum height will be:

=ucosθ

=ucos37°

=10 x 4/5

=2 x 4

=8 m/sec.

Hence, the speed at the maximum height is 8 m/sec.

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