Question

14. A player throws a ball at an initial velocity of 36 m/second. The maximum
distance the ball can reach (assume ball is caught at the same height at
which it was released) is:
a. 146 m
b. 130 m
c.
132 m
d. 129 m
e. 145 m
​

Answer 1

Answer:

maximum distance = range

as we know

R = V^2 sin2 theta/ g

v = 36m/s

= 36^2 sin 2(45) as range is maximum at 45 degree

= 1296/9.8

=132.2m = 132m approx

Answer 2

Answer:

Explanation:

A particle when thrown upward experiences two types of forces that point opposite to the particle's motion. These forces are the weight of the ball and the air resistance. In situations where air resistance is neglected, the ball would then accelerate due to gravity equal to g =9.8 m/s2

Momentarily, this particle would then have a velocity equal to zero when it reaches the maximum height then goes downwards into a free-fall motion.

Horizontal Range of Projectile:

If we throw an object at any angle with the horizontal with some initial velocity and let it to move freely only under the effect of earth's gravity, we call the object projectile.

Let a projectile be projected with initial velocity of magnitude

v0 at an angle α with the horizontal.

Horizontal range, i.e. maximum horizontal displacement of the projectile is given by:

.$R = v^{2} /g(sin 2\alpha )$

If magnitude of initial velocity remains unchanged, maximum horizontal range will be obtained by putting

$2\alpha = 90$°

and it is given by:

$R(max)= v^2/g$

thus putting values in equation we get:

$36^2/9.8\\=1296/9.8\\=132.2448$

Rounding off we get 132m

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