A ball thrown with a speed of 100ms^-1 attains a height of 150m. ( take g= 9.8ms^-2) calculate the: (a) time of flight (b) angle of projection (c) range?
Recall the basic equation of motion here. The height y is
y = 100sinθ t - 4.9t^2
and the ball's upward speed is
v(t) = 100sinθ - 9.8t
(b) The max height is achieved when the velocity becomes zero. That is, when
t = 100sinθ/9.8 = 10.2 sinθ
Now, we know that the max height is 150, so
100sinθ * 10.2 - 4.9*10.2^2 = 150
θ = 40°
(a) Now we know that our height is roughly
y = 64.7t - 4.9t^2
So, the time of flight is 13.2 seconds
(c) since the horizontal speed is a constant 100cosθ = 76.6 m/s, the range is
76.6 * 13.2 = 1011 m
Y^2 = Yo^2 + 2g*h = 0,
Yo^2 + (-19.6)150 = 0.
Yo = 54.2 m/s. = Ver. component of initial velocity.
b. Yo = Vo*sinA = 54.2,
100*sinA = 54.2,
A = 32.8o.
c. Range = Vo^2*sin(2A)/g = 100^2*sin(65.6)/9.8 = 929.3 m.
a. Range = Xo*T = 100*Cos32.8 * T = 929.3,
T = 11.1 s. = Time of flight.
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To find the answers to the questions, we can use the equations of projectile motion. Let's break down each part:
(a) Time of flight:
The time of flight is the total time it takes for the ball to reach its maximum height, then come back down and hit the ground. In this case, we need to find the time it takes for the ball to reach a height of 150m.
We can use the equation for vertical displacement in projectile motion:
Δy = v₀y * t + (1/2) * g * t²
Where:
Δy is the vertical displacement (150m in this case),
v₀y is the initial vertical velocity (which is equal to the initial speed multiplied by the sine of the angle of projection),
g is the acceleration due to gravity (9.8m/s² in this case), and
t is the time of flight that we are trying to find.
Since the ball is thrown vertically upwards, the initial vertical velocity is equal to the initial speed (100m/s) * sine(angle of projection).
Let's plug in the values we know and solve for t:
150 = (100 * sin(angle)) * t - (1/2) * 9.8 * t²
Now we can solve this quadratic equation for t. Rearranging the equation, we get:
4.9t² - 100sin(angle)t + 150 = 0
By solving this equation, we can find the time of flight.
(b) Angle of projection:
The angle of projection is the angle at which the ball is thrown with respect to the horizontal direction. We can calculate this angle using the equation for vertical velocity:
v₀y = v * sin(angle)
Where:
v₀y is the initial vertical velocity (which we already know as 100m/s),
v is the initial speed (100m/s), and
angle is the angle of projection.
Rearranging the equation, we get:
sin(angle) = v₀y / v
By taking the inverse sine (sin⁻¹) of both sides, we can find the angle of projection.
(c) Range:
The range is the horizontal distance covered by the ball. We can calculate the range using the equation for horizontal displacement:
Δx = v₀x * t
Where:
Δx is the horizontal displacement or range,
v₀x is the initial horizontal velocity (which is equal to the initial speed multiplied by the cosine of the angle of projection), and
t is the time of flight that we found in part (a).
Since the ball was thrown vertically upwards, the initial horizontal velocity is equal to the initial speed (100m/s) multiplied by the cosine of the angle of projection.
Let's plug in the values we know and solve for Δx:
Δx = (100 * cos(angle)) * t
By substituting the value of t we found in part (a), we can calculate the range.